Prove the following: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x

Prove the following: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1

Class 11 | Exercise 3.3 | Maths | NCERT | Trigonometric Functions

Prove the following: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1

ACCORDING TO THE QUES,

$L.H.S.\text{ }=\text{ }cos\text{ }6x$

OR,

$=\text{ }cos\text{ }3\left( 2x \right)$

FORMULA SAYS,

$~cos\text{ }3A\text{ }=\text{ }4\text{ }cos3\text{ }A\text{ }\text{ }3\text{ }cos\text{ }A$

$=\text{ }4\text{ }cos3\text{ }2x\text{ }\text{ }3\text{ }cos\text{ }2x$

NOW,

$cos\text{ }2x\text{ }=\text{ }2\text{ }cos2\text{ }x\text{ }\text{ }1$

(according to the formula)

$=\text{ }4\text{ }[\left( 2\text{ }cos2\text{ }x\text{ }\text{ }1 \right)3\text{ }\text{ }3\text{ }\left( 2\text{ }cos2\text{ }x\text{ }\text{ }1 \right)$

OR,

$=\text{ }4\text{ }\left[ \left( 2\text{ }cos2\text{ }x \right)\text{ }3\text{ }\text{ }\left( 1 \right)3\text{ }\text{ }3\text{ }\left( 2\text{ }cos2\text{ }x \right)\text{ }2\text{ }+\text{ }3\text{ }\left( 2\text{ }cos2\text{ }x \right) \right]\text{ }\text{ }6cos2\text{ }x\text{ }+\text{ }3$

WE HAVE,

$=\text{ }4\text{ }\left[ 8cos6x\text{ }\text{ }1\text{ }\text{ }12\text{ }cos4x\text{ }+\text{ }6\text{ }cos2x \right]\text{ }\text{ }6\text{ }cos2x\text{ }+\text{ }3$

OR,

$=\text{ }32\text{ }cos6x\text{ }\text{ }4\text{ }\text{ }48\text{ }cos4x\text{ }+\text{ }24\text{ }cos2\text{ }x\text{ }\text{ }6\text{ }cos2x\text{ }+\text{ }3$

OR,

$=\text{ }32\text{ }cos6x\text{ }\text{ }48\text{ }cos4x\text{ }+\text{ }18\text{ }cos2x\text{ }\text{ }1$

$=\text{ }R.H.S.$

Find the vector equation of the line passing through the point with position vector and parallel to the vector Deduce the Cartesian equations of the line.

A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

A line passes through the point (3, 4, 5) and is parallel to the vector Find the equations of the line in the vector as well as Cartesian forms.

Evaluate:

If $f(x) = |x| - 3,find\mathop {\lim }\limits_{x \to 3} f(x)$

Let $\begin{array}{l} f(x) = \left\{ \begin{array}{l} \frac{x}{{|x|}},x \ne 0\\ 0,x = 0 \end{array} \right.\\ \end{array}$ . Show that $\begin{array}{l} \mathop {\lim }\limits_{x \to 0} f(x) \end{array}$ does not exist.

Let $\begin{array}{l} f(x) = \left\{ \begin{array}{l} \frac{|x-3|}{{(x-3)'}},x \ne 3\\ 0,x = 3 \end{array} \right.\\ \end{array}$ . Show that $\begin{array}{l} \mathop {\lim }\limits_{x \to 3} f(x) \end{array}$ does not exist.

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If $f(x) = |x| - 3,find\mathop {\lim }\limits_{x \to 3} f(x)$

Let $\begin{array}{l} f(x) = \left\{ \begin{array}{l} \frac{x}{{|x|}},x \ne 0\\ 0,x = 0 \end{array} \right.\\ \end{array}$ . Show that $\begin{array}{l} \mathop {\lim }\limits_{x \to 0} f(x) \end{array}$ does not exist.

Let $\begin{array}{l} f(x) = \left\{ \begin{array}{l} \frac{|x-3|}{{(x-3)'}},x \ne 3\\ 0,x = 3 \end{array} \right.\\ \end{array}$ . Show that $\begin{array}{l} \mathop {\lim }\limits_{x \to 3} f(x) \end{array}$ does not exist.

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