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Prove that: \[\mathbf{sin}\text{ }\mathbf{5x}\text{ }=\text{ }\mathbf{5}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x}\text{ }\mathbf{sin}\text{ }\mathbf{x}\text{ }\text{ }\mathbf{10}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{x}\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{3}}}~\mathbf{x}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{5}}}~\mathbf{x}\]
Prove that: \[\mathbf{sin}\text{ }\mathbf{5x}\text{ }=\text{ }\mathbf{5}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x}\text{ }\mathbf{sin}\text{ }\mathbf{x}\text{ }\text{ }\mathbf{10}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{x}\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{3}}}~\mathbf{x}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{5}}}~\mathbf{x}\]
CBSE | Class 11 | Exercise 9.2 | Maths | RD Sharma | Trigonometric Functions at Multiples and Submultiples of an Angle
Prove that: \[\mathbf{sin}\text{ }\mathbf{5x}\text{ }=\text{ }\mathbf{5}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{4}}}~\mathbf{x}\text{ }\mathbf{sin}\text{ }\mathbf{x}\text{ }\text{ }\mathbf{10}\text{ }\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}~\mathbf{x}\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{3}}}~\mathbf{x}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{5}}}~\mathbf{x}\]

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,

\[Sin\text{ }\left( x\text{ }+\text{ }y \right)\text{ }=\text{ }sin\text{ }x\text{ }cos\text{ }y\text{ }+\text{ }cos\text{ }x\text{ }sin\text{ }y\ldots ..\left( i \right)\]

So,

\[sin\text{ }5x\text{ }=\text{ }sin\text{ }3x\text{ }cos\text{ }2x\text{ }+\text{ }cos\text{ }3x\text{ }sin\text{ }2x\]

\[=\text{ }sin\text{ }\left( 2x\text{ }+\text{ }x \right)\text{ }cos\text{ }2x\text{ }+\text{ }cos\text{ }\left( 2x\text{ }+\text{ }x \right)\text{ }sin\text{ }2x\] ……..(ii)

And

cos (x + y) = cos x cos y – sin x sin y……(iii)

Now substituting equation (i) and (iii) in equation (ii), we get

\[sin\text{ }5x\text{ }=\text{ }\left( sin\text{ }2x\text{ }cos\text{ }x\text{ }+\text{ }cos\text{ }2x\text{ }sin\text{ }x\text{ } \right)\text{ }cos\text{ }2x\text{ }+\text{ }\left( \text{ }cos\text{ }2x\text{ }cos\text{ }x\text{ }\text{ }sin\text{ }2x\text{ }sin\text{ }x \right)\text{ }sin\text{ }2x\]… (iv)

Now \[sin\text{ }2x\text{ }=\text{ }2sin\text{ }x\text{ }cos\text{ }x\] ………(v)

And \[cos\text{ }2x\text{ }=\text{ }co{{s}^{2}}x\text{ }\text{ }si{{n}^{2}}x\] ………(vi)

Substituting equation (v) and (vi) in equation (iv), we get

\[sin\text{ }5x\text{ }=\text{ }[\left( 2\text{ }sin\text{ }x\text{ }cos\text{ }x \right)\text{ }cos\text{ }x\text{ }+\text{ }(co{{s}^{2}}x\text{ }\text{ }si{{n}^{2}}x)\text{ }sin\text{ }x]\text{ }(co{{s}^{2}}x\text{ }\text{ }si{{n}^{2}}x)\text{ }+\text{ }[(co{{s}^{2}}x\text{ }\text{ }si{{n}^{2}}x)\text{ }cos\text{ }x\text{ }\text{ }\left( 2\text{ }sin\text{ }x\text{ }cos\text{ }x \right)\text{ }sin\text{ }x)]\text{ }\left( 2\text{ }sin\text{ }x\text{ }cos\text{ }x \right)\]

\[=\text{ }[2\text{ }sin\text{ }x\text{ }co{{s}^{2}}~x\text{ }+\text{ }sin\text{ }x\text{ }co{{s}^{2}}x\text{ }\text{ }si{{n}^{3}}x]\text{ }(co{{s}^{2}}x\text{ }\text{ }si{{n}^{2}}x)\text{ }+\text{ }[co{{s}^{3}}x\text{ }\text{ }si{{n}^{2}}x\text{ }cos\text{ }x\text{ }\text{ }2\text{ }si{{n}^{2}}~x\text{ }cos\text{ }x]\text{ }\left( 2\text{ }sin\text{ }x\text{ }cos\text{ }x \right)\]

\[=\text{ }co{{s}^{2}}x\text{ }[3\text{ }sin\text{ }x\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{3}}x]\text{ }\text{ }si{{n}^{2}}x\text{ }[3\text{ }sin\text{ }x\text{ }co{{s}^{2}}~x\text{ }\text{ }si{{n}^{3}}x]\text{ }+\text{ }2\text{ }sin\text{ }x\text{ }co{{s}^{4}}x\text{ }\text{ }2\text{ }si{{n}^{3}}~x\text{ }co{{s}^{2}}~x\text{ }\text{ }4\text{ }si{{n}^{3}}~x\text{ }co{{s}^{2}}~x\]

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

= RHS

Hence proved.

i

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