# Find $a \cos 2\alpha$ $+$ $b\sin 2\alpha$ if $\dfrac{\cos \alpha}{a} = \dfrac{\sin \alpha}{b}$

$\alpha$ is an angle of a right angle triangle. The relation between trigonometric functions $\cos \alpha$ and $\sin \alpha$ is expressed in a ratio in terms of literal numbers $a$ and $b$.

The relation is expressed in mathematical form as $\dfrac{\cos \alpha}{a} = \dfrac{\sin \alpha}{b}$

It is useful to express $\cos \alpha$ in terms of $\sin \alpha$ and vice versa in this problem to find the value of $a \cos 2\alpha + b\sin 2\alpha$

The problem can be solved in trigonometry in two methods by using relation between both trigonometric functions.

### Method: 1

###### Step: 1

According to double angle trigonometric identities,

$(1) \,\,\,\,\,\,$ $\cos 2\alpha = 2\cos^2 \alpha -1$

$(2) \,\,\,\,\,\,$ $\sin 2\alpha = 2\sin \alpha \cos \alpha$

Now, expand $\cos 2\alpha$ and $\sin 2\alpha$ in the trigonometric expression.

$= a \times (2\cos^2 \alpha -1)$ $+$ $b \times (2\sin \alpha \cos \alpha)$

$= 2a\cos^2 \alpha -a$ $+$ $2b\sin \alpha \cos \alpha$

###### Step: 2

If $\dfrac{\cos \alpha}{a} = \dfrac{\sin \alpha}{b}$, then $\cos \alpha = \Bigg[\dfrac{a}{b}\Bigg]\sin \alpha$. Replace $\cos \alpha$ by its value in the second term of the expression.

$= 2a\cos^2 \alpha -a$ $+$ $2b\sin \alpha \times \Bigg[\dfrac{a}{b}\Bigg]\sin \alpha$

$= 2a\cos^2 \alpha -a$ $+$ $2b\Bigg[\dfrac{a}{b}\Bigg]\sin \alpha \sin \alpha$

$= 2a\cos^2 \alpha -a$ $+$ $2a\Bigg[\dfrac{b}{b}\Bigg]\sin^2 \alpha$

$= 2a\cos^2 \alpha -a$ $+$ $\require{cancel} 2a\Bigg[\dfrac{\cancel{b}}{\cancel{b}}\Bigg]\sin^2 \alpha$

$= 2a\cos^2 \alpha -a$ $+$ $2a\sin^2 \alpha$

$= 2a\cos^2 \alpha + 2a\sin^2 \alpha -a$

###### Step: 3

Take $2a$ common and proceed to the simplification.

$= 2a(\cos^2 \alpha + \sin^2 \alpha)$ $-a$

According to Pythagorean trigonometric identity, the sum of squares of sine and cosine of an angle is one.

$= 2a(1)$ $-a$

$= 2a-a$

$\therefore \,\,\,\,\,\, a \cos 2\alpha$ $+$ $b\sin 2\alpha = a$

It is the required solution for this trigonometric problem.

### Method: 2

###### Step: 1

As per the double angle trigonometric identities,

$(1) \,\,\,\,\,\,$ $\cos 2\alpha = 1-2\sin^2 \alpha$

$(2) \,\,\,\,\,\,$ $\sin 2\alpha = 2\sin \alpha \cos \alpha$

Expand sine and cosine of double angle and replace them in the trigonometric expression.

$= a(1-2\sin^2 \alpha) + b(2 \sin \alpha \cos \alpha)$

$= a(1-2\sin^2 \alpha) + 2b \sin \alpha \cos \alpha$

$= a(1-2\sin^2 \alpha) + 2\sin \alpha (b\cos \alpha)$

###### Step: 2

If $\dfrac{\cos \alpha}{a} = \dfrac{\sin \alpha}{b}$, then $b\cos \alpha = a\sin \alpha$.

$= a(1-2\sin^2 \alpha) + 2\sin \alpha (a\sin \alpha)$

$= a-2a\sin^2 \alpha + 2a\sin \alpha \times \sin \alpha)$

$= a-2a\sin^2 \alpha + 2a\sin^2 \alpha$

$\require{cancel} = a-\cancel{2a\sin^2 \alpha} + \cancel{2a\sin^2 \alpha}$

$\therefore \,\,\,\,\,\, a \cos 2\alpha$ $+$ $b\sin 2\alpha = a$

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