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CBSE Class 11 RD Sharma Maths Trigonometric Functions at Multiples and Submultiples of an Angle
Trigonometric Functions at Multiples and Submultiples of an Angle

Prove that: \[\mathbf{cos}\text{ }\mathbf{\pi }/\mathbf{15}\text{ }\mathbf{cos}\text{ }\mathbf{2\pi }/\mathbf{15}\text{ }\mathbf{cos}\text{ }\mathbf{4\pi }/\mathbf{15}\text{ }\mathbf{cos}\text{ }\mathbf{7\pi }/\mathbf{15}\text{ }=\text{ }\mathbf{1}/\mathbf{16}\]

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Let us consider LHS: \[\mathbf{cos}\text{ }\mathbf{\pi }/\mathbf{15}\text{ }\mathbf{cos}\text{ }\mathbf{2\pi }/\mathbf{15}\text{ }\mathbf{cos}\text{ }\mathbf{4\pi }/\mathbf{15}\text{...

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Prove that: \[\mathbf{cos}\text{ }\mathbf{7}{{\mathbf{8}}^{\mathbf{o}}}~\mathbf{cos}\text{ }\mathbf{4}{{\mathbf{2}}^{\mathbf{o}}}~\mathbf{cos}\text{ }\mathbf{3}{{\mathbf{6}}^{\mathbf{o}}}~=\text{ }\mathbf{1}/\mathbf{8}\]

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Let us consider LHS: \[cos\text{ }{{78}^{o}}~cos\text{ }{{42}^{o}}~cos\text{ }{{36}^{o}}\] Let us multiply and divide by 2 we get, \[cos\text{ }{{78}^{o}}~cos\text{ }{{42}^{o}}~cos\text{...

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Prove that: \[\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~\mathbf{4}{{\mathbf{2}}^{\mathbf{o}}}~\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{7}{{\mathbf{8}}^{\mathbf{o}}}~=\text{ }\left( \surd \mathbf{5}\text{ }+\text{ }\mathbf{1} \right)/\mathbf{8}\]

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Let us consider LHS: \[si{{n}^{2}}~{{42}^{o}}~\text{ }co{{s}^{2}}~{{78}^{o}}~=\text{ }si{{n}^{2}}~({{90}^{o}}~\text{ }{{48}^{o}})\text{ }\text{ }co{{s}^{2}}~({{90}^{o}}~\text{ }{{12}^{o}})\]...

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Prove that: \[\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{4}}^{\mathbf{o}}}~\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~{{\mathbf{6}}^{\mathbf{o}}}~=\text{ }\left( \surd \mathbf{5}\text{ }\text{ }\mathbf{1} \right)/\mathbf{8}\]

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Let us consider LHS: \[si{{n}^{2}}~{{24}^{o}}~\text{ }si{{n}^{2}}~{{6}^{o}}\] we know, \[sin\text{ }\left( A\text{ }+\text{ }B \right)\text{ }sin\text{ }\left( A\text{ }\text{ }B \right)\text{...

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Prove that: \[~\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~\mathbf{2\pi }/\mathbf{5}\text{ }\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~\mathbf{\pi }/\mathbf{3}\text{ }=\text{ }\left( \surd \mathbf{5}\text{ }\text{ }\mathbf{1} \right)/\mathbf{8}\]

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Let us consider LHS: \[si{{n}^{2}}~2\pi /5\text{ }\text{ }si{{n}^{2}}~\pi /3\text{ }=\text{ }si{{n}^{2}}~\left( \pi /2\text{ }\text{ }\pi /10 \right)\text{ }\text{ }si{{n}^{2}}~\pi /3\] we know,...

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Prove that: \[\mathbf{sin}\text{ }\mathbf{5x}\text{ }=\text{ }\mathbf{5}\text{ }\mathbf{sin}\text{ }\mathbf{x}\text{ }\text{ }\mathbf{20}\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{3}}}~\mathbf{x}\text{ }+\text{ }\mathbf{16}\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{5}}}~\mathbf{x}\]

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Let us consider LHS: \[sin\text{ }5x\] We can write \[sin\text{ }5x\] as, \[sin\text{ }5x\text{ }=\text{ }sin\text{ }\left( 3x\text{ }+\text{ }2x \right)\] From formula, \[Sin\text{ }\left( x\text{...

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Prove that: \[\mathbf{sin}\text{ }\mathbf{5x}\text{ }=\text{ }\mathbf{5}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x}\text{ }\mathbf{sin}\text{ }\mathbf{x}\text{ }\text{ }\mathbf{10}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{x}\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{3}}}~\mathbf{x}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{5}}}~\mathbf{x}\]

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Let us consider LHS: \[sin\text{ }5x\] We can write \[sin\text{ }5x\] as, \[sin\text{ }5x\text{ }=\text{ }sin\text{ }\left( 3x\text{ }+\text{ }2x \right)\] From formula, \[Sin\text{ }\left( x\text{...

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Prove that: \[\mathbf{co}{{\mathbf{s}}^{\mathbf{3}}}~\mathbf{x}\text{ }\mathbf{sin}\text{ }\mathbf{3x}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{3}}}~\mathbf{x}\text{ }\mathbf{cos}\text{ }\mathbf{3x}\text{ }=\text{ }\mathbf{3}/\mathbf{4}\text{ }\mathbf{sin}\text{ }\mathbf{4x}\]

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We know that, \[cos\text{ }3\theta \text{ }=\text{ }4co{{s}^{3}}\theta \text{ }\text{ }3cos\theta \] SO, \[4\text{ }co{{s}^{3}}\theta \text{ }=\text{ }cos3\theta \text{ }+\text{ }3cos\theta \]...

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Prove that: \[\mathbf{4}\text{ }(\mathbf{co}{{\mathbf{s}}^{\mathbf{3}}}~\mathbf{1}{{\mathbf{0}}^{\mathbf{o}}}~+\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{3}}}~\mathbf{2}{{\mathbf{0}}^{\mathbf{o}}})\text{ }=\text{ }\mathbf{3}\text{ }(\mathbf{cos}\text{ }\mathbf{1}{{\mathbf{0}}^{\mathbf{o}}}~+\text{ }\mathbf{sin}\text{ }\mathbf{2}{{\mathbf{0}}^{\mathbf{o}}})\]

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Take LHS: \[4\text{ }(co{{s}^{3}}~{{10}^{o}}~+\text{ }si{{n}^{3}}~{{20}^{o}})\] We know that, \[sin\text{ }{{60}^{o}}~=~\sqrt{3/2}\text{ }=\text{ }cos\text{ }{{30}^{o}}\] \[Sin\text{...

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Prove that: \[\mathbf{sin}\text{ }\mathbf{5x}\text{ }=\text{ }\mathbf{5}\text{ }\mathbf{sin}\text{ }\mathbf{x}\text{ }\text{ }\mathbf{20}\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{3}}}~\mathbf{x}\text{ }+\text{ }\mathbf{16}\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{5}}}~\mathbf{x}\]

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Take LHS: \[sin\text{ }5x\] We can write \[sin\text{ }5x\] as, \[sin\text{ }5x\text{ }=\text{ }sin\text{ }\left( 3x\text{ }+\text{ }2x \right)\] From formula, \[Sin\text{ }\left( x\text{ }+\text{ }y...

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Prove the following identities: \[\mathbf{tan}\text{ }\left( \mathbf{\pi }/\mathbf{4}\text{ }+\text{ }\mathbf{x} \right)\text{ }+\text{ }\mathbf{tan}\text{ }\left( \mathbf{\pi }/\mathbf{4}\text{ }\text{ }\mathbf{x} \right)\text{ }=\text{ }\mathbf{2}\text{ }\mathbf{sec}\text{ }\mathbf{2x}\]

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Take LHS: \[tan\text{ }\left( \pi /4\text{ }+\text{ }x \right)\text{ }+\text{ }tan\text{ }\left( \pi /4\text{ }\text{ }x \right)\] From formula, \[tan\text{ }\left( A+B \right)\text{ }=\text{...

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Prove the following identities: \[\mathbf{co}{{\mathbf{s}}^{\mathbf{6}}}~\mathbf{x}\text{ }\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{6}}}~\mathbf{x}\text{ }=\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }(\mathbf{1}\text{ }\text{ }\mathbf{1}/\mathbf{4}\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~\mathbf{2x})\]

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Take LHS: \[co{{s}^{6}}~x\text{ }\text{ }si{{n}^{6}}~x\] From formula, \[{{\left( a\text{ }+\text{ }b \right)}^{~2}}~=\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}~+\text{ }2ab\] \[{{a}^{3}}~\text{...

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Prove the following identities: \[\mathbf{2}(\mathbf{si}{{\mathbf{n}}^{\mathbf{6}}}~\mathbf{x}\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{6}}}~\mathbf{x})\text{ }\text{ }\mathbf{3}(\mathbf{si}{{\mathbf{n}}^{\mathbf{4}}}~\mathbf{x}\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x})\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}\]

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Take LHS: \[2(si{{n}^{6}}~x\text{ }+\text{ }co{{s}^{6}}~x)\text{ }\text{ }3(si{{n}^{4}}~x\text{ }+\text{ }co{{s}^{4}}~x)\text{ }+\text{ }1\] From formula, \[{{\left( a\text{ }+\text{ }b...

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Prove the following identities: \[\mathbf{3}{{\left( \mathbf{sin}\text{ }\mathbf{x}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{x} \right)}^{~\mathbf{4}}}~+\text{ }\mathbf{6}\text{ }{{\left( \mathbf{sin}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{x} \right)}^{~\mathbf{2}}}~+\text{ }\mathbf{4}\text{ }(\mathbf{si}{{\mathbf{n}}^{\mathbf{6}}}~\mathbf{x}\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{6}}}~\mathbf{x})\text{ }=\text{ }\mathbf{13}\]

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Take LHS: \[3{{\left( sin\text{ }x\text{ }\text{ }cos\text{ }x \right)}^{~4}}~+\text{ }6\text{ }{{\left( sin\text{ }x\text{ }+\text{ }cos\text{ }x \right)}^{~2}}~+\text{ }4\text{...

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Prove the following identities: \[\mathbf{sin}\text{ }\mathbf{4x}\text{ }=\text{ }\mathbf{4}\text{ }\mathbf{sin}\text{ }\mathbf{x}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{3}}}~\mathbf{x}\text{ }\text{ }\mathbf{4}\text{ }\mathbf{cos}\text{ }\mathbf{x}\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{3}}}~\mathbf{x}\]

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Take LHS: \[sin\text{ }4x\] From formula, \[sin\text{ 2}x\text{ }=\text{ }2\text{ }sin\text{ }x\text{ }cos\text{ }x\] \[cos\text{ }2x\text{ }=\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x\] So,...

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Prove the following identities: \[\mathbf{cos}\text{ }\mathbf{4x}\text{ }=\text{ }\mathbf{1}\text{ }\text{ }\mathbf{8}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{x}\text{ }+\text{ }\mathbf{8}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x}\]

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Take LHS: \[cos\text{ }4x\] From formula, \[\text{ }cos\text{ }2x\text{ }=\text{ }2\text{ }co{{s}^{2}}~x\text{ }\text{ }1\] So, \[cos\text{ }4x\text{ }=\text{ }2\text{ }co{{s}^{2}}~2x\text{ }\text{...

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Prove the following identities: \[\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\left( \mathbf{\pi }/\mathbf{4}\text{ }\text{ }\mathbf{x} \right)\text{ }\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~\left( \mathbf{\pi }/\mathbf{4}\text{ }\text{ }\mathbf{x} \right)\text{ }=\text{ }\mathbf{sin}\text{ }\mathbf{2x}\]

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Take LHS: \[co{{s}^{2}}~\left( \pi /4\text{ }\text{ }x \right)\text{ }\text{ }si{{n}^{2}}~\left( \pi /4\text{ }\text{ }x \right)\] From formula, \[co{{s}^{2}}~A\text{ }\text{ }si{{n}^{2}}~A\text{...

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Prove the following identities: \[\left( \mathbf{sin}\text{ }\mathbf{3x}\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{x} \right)\text{ }\mathbf{sin}\text{ }\mathbf{x}\text{ }+\text{ }\left( \mathbf{cos}\text{ }\mathbf{3x}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{x} \right)\text{ }\mathbf{cos}\text{ }\mathbf{x}\text{ }=\text{ }\mathbf{0}\]

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LHS: \[\left( sin\text{ }3x\text{ }+\text{ }sin\text{ }x \right)\text{ }sin\text{ }x\text{ }+\text{ }\left( cos\text{ }3x\text{ }\text{ }cos\text{ }x \right)\text{ }cos\text{ }x\] \[=\text{ }\left(...

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Prove the following identities: \[\mathbf{co}{{\mathbf{s}}^{\mathbf{3}}}~\mathbf{2x}\text{ }+\text{ }\mathbf{3}\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }=\text{ }\mathbf{4}\text{ }(\mathbf{co}{{\mathbf{s}}^{\mathbf{6}}}~\mathbf{x}\text{ }\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{6}}}~\mathbf{x})\]

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Take RHS: \[4\text{ }(co{{s}^{6}}~x\text{ }\text{ }si{{n}^{6}}~x)\] By expanding we get, \[4\text{ }(co{{s}^{6}}~x\text{ }\text{ }si{{n}^{6}}~x)\text{ }=\text{ }4\text{...

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Prove the following identities: \[\mathbf{1}\text{ }+\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{2x}\text{ }=\text{ }\mathbf{2}\text{ }(\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{4}}}~\mathbf{x})\]

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Take LHS: \[1\text{ }+\text{ }co{{s}^{2}}~2x\] From formula, \[cos2x\text{ }=\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x\] \[co{{s}^{2}}~x\text{ }+\text{ }si{{n}^{2}}~x\text{ }=\text{ }1\]...

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Prove the following identities: \[\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~\left( \mathbf{\pi }/\mathbf{8}\text{ }+\text{ }\mathbf{x}/\mathbf{2} \right)\text{ }\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~\left( \mathbf{\pi }/\mathbf{8}\text{ }\text{ }\mathbf{x}/\mathbf{2} \right)\text{ }=\text{ }\mathbf{1}/\surd \mathbf{2}\text{ }\mathbf{sin}\text{ }\mathbf{x}\]

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  Take LHS: \[si{{n}^{2}}~\left( \pi /8\text{ }+\text{ }x/2 \right)\text{ }\text{ }si{{n}^{2}}~\left( \pi /8\text{ }\text{ }x/2 \right)\] From formula, \[\text{ }si{{n}^{2}}~A\text{ }\text{...

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Prove the following identities: \[{{\left( \mathbf{cos}\text{ }\mathbf{\alpha }\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{\beta } \right)}^{~\mathbf{2}}}~+\text{ }{{\left( \mathbf{sin}\text{ }\mathbf{\alpha }\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{\beta } \right)}^{~\mathbf{2}}}~=\text{ }\mathbf{4}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\left( \mathbf{\alpha }\text{ }\text{ }\mathbf{\beta } \right)/\mathbf{2}\]

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  Take LHS: \[{{\left( cos\text{ }\alpha \text{ }+\text{ }cos\text{ }\beta  \right)}^{2}}~+\text{ }{{\left( sin\text{ }\alpha \text{ }+\text{ }sin\text{ }\beta  \right)}^{2}}\] By expanding we...

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Prove the following identities: \[\mathbf{cos}\text{ }\mathbf{x}\text{ }/\text{ }\left( \mathbf{1}\text{ }\text{ }\mathbf{sin}\text{ }\mathbf{x} \right)\text{ }=\text{ }\mathbf{tan}\text{ }\left( \mathbf{\pi }/\mathbf{4}\text{ }+\text{ }\mathbf{x}/\mathbf{2} \right)\]

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Take LHS: \[cos\text{ }x\text{ }/\text{ }\left( 1\text{ }\text{ }sin\text{ }x \right)\] From Formula, \[cos\text{ }2x\text{ }=\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{2}}~x\] \[Cos\text{...

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Prove the following identities: \[\mathbf{cos}\text{ }\mathbf{2x}\text{ }/\text{ }\left( \mathbf{1}\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{2x} \right)\text{ }=\text{ }\mathbf{tan}\text{ }\left( \mathbf{\pi }/\mathbf{4}\text{ }\text{ }\mathbf{x} \right)\]

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Take LHS: \[\begin{align} & cos\text{ }2x\text{ }/\text{ }\left( 1\text{ }+\text{ }sin\text{ }2x \right) \\ &  \\ &  \\ &  \\ &  \\ \end{align}\] From Formula, \[cos\text{...

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Prove the following identities: \[\left[ \mathbf{sin}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{2x} \right]\text{ }/\text{ }\left[ \mathbf{1}\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{2x} \right]\text{ }=\text{ }\mathbf{tan}\text{ }\mathbf{x}\]

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Take LHS: \[\left[ sin\text{ }x\text{ }+\text{ }sin\text{ }2x \right]\text{ }/\text{ }\left[ 1\text{ }+\text{ }cos\text{ }x\text{ }+\text{ }cos\text{ }2x \right]\] From formulae, \[cos\text{...

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Prove the following identities: \[.\text{ }\left[ \mathbf{1}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{2x} \right]\text{ }/\text{ }\left[ \mathbf{1}\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{sin}\text{ }\mathbf{2x} \right]\text{ }=\text{ }\mathbf{tan}\text{ }\mathbf{x}\]

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Take LHS: \[\left[ 1\text{ }\text{ }cos\text{ }2x\text{ }+\text{ }sin\text{ }2x \right]\text{ }/\text{ }\left[ 1\text{ }+\text{ }cos\text{ }2x\text{ }+\text{ }sin\text{ }2x \right]\] From formulae,...

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Prove the following identities: \[\mathbf{sin}\text{ }\mathbf{2x}\text{ }/\text{ }\left( \mathbf{1}\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{2x} \right)\text{ }=\text{ }\mathbf{tan}\text{ }\mathbf{x}\]

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Take LHS: \[sin\text{ }2x\text{ }/\text{ }\left( 1\text{ }+\text{ }cos\text{ }2x \right)\] From formulae, \[cos\text{ }2x\text{ }=\text{ }1\text{ }\text{ }2\text{ }si{{n}^{2}}~x\] \[=\text{ }2\text{...

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Prove the following identities: \[\mathbf{sin}\text{ }\mathbf{2x}\text{ }/\text{ }\left( \mathbf{1}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{2x} \right)\text{ }=\text{ }\mathbf{cot}\text{ }\mathbf{x}\]

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Take LHS: \[sin\text{ }2x\text{ }/\text{ }\left( 1\text{ }\text{ }cos\text{ }2x \right)\] From Formula, \[\text{ }cos\text{ }2x\text{ }=\text{ }1\text{ }\text{ }2\text{ }si{{n}^{2}}~x\], \[Sin\text{...

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Prove the following identities: \[\sqrt{\left[ \left( \mathbf{1}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{2x} \right)\text{ }/\text{ }\left( \mathbf{1}\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{2x} \right) \right]\text{ }}=\text{ }\mathbf{tan}\text{ }\mathbf{x}\]

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Let us consider LHS: \[\sqrt{\left[ \left( \mathbf{1}\text{ }\text{ }\mathbf{cos}\text{ }\mathbf{2x} \right)\text{ }/\text{ }\left( \mathbf{1}\text{ }+\text{ }\mathbf{cos}\text{ }\mathbf{2x} \right)...

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