Question

(a) Show that $\cos 3\theta - \sin 3\theta = (\cos \theta + \sin \theta )(1 - 2\sin 2\theta )$
(b) If $\tan A = \dfrac{5}{6}$ and $\tan B = \dfrac{1}{{11}}$ then show that $A + B = \dfrac{\pi }{4}$ or $\dfrac{{5\pi }}{4}$

Answer 1

Here, we have to solve trigonometric functions. In order to solve this questions we first consider left side of the equation and then solve it and prove equals to the right side of the equation by using various trigonometric identities such as $\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta$, $\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta$ and $\tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$.

Complete step by step answer:
Here, we have to solve trigonometric functions.
(a) we have $\cos 3\theta - \sin 3\theta = (\cos \theta + \sin \theta )(1 - 2\sin 2\theta )$
Consider left side of the equation i.e., $\cos 3\theta - \sin 3\theta$
We know that $\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta$ and $\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta$
Substituting these values, we get,
$\Rightarrow \cos 3\theta - \sin 3\theta = 4{\cos ^3}\theta - 3\cos \theta - (3\sin \theta - 4{\sin ^3}\theta )$
Rearranging the above equation. We get,
$\Rightarrow 4{\cos ^3}\theta + 4{\sin ^3}\theta - 3\cos \theta - 3\sin \theta$

The above equation can be written as
$\Rightarrow 4({\cos ^3}\theta + {\sin ^3}\theta ) - 3(\cos \theta + \sin \theta )$
Now using the formula ${(a + b)^3} = (a + b)({a^2} - ab + {b^2})$. We get,
$\Rightarrow 4(\cos \theta + \sin \theta )({\cos ^2}\theta - \cos \theta \sin \theta + {\sin ^2}\theta ) - 3(\cos \theta + \sin \theta )$
We know that ${\cos ^2}\theta + {\sin ^2}\theta = 1$. So,
$\Rightarrow 4(\cos \theta + \sin \theta )(1 - \cos \theta \sin \theta ) - 3(\cos \theta + \sin \theta )$

Taking common $(\cos \theta + \sin \theta )$ from the above equation. We get,
$\Rightarrow (\cos \theta + \sin \theta )[4(1 - \cos \theta \sin \theta ) - 3]$
Solving the above equation. We get,
$\Rightarrow (\cos \theta + \sin \theta )(4 - 4\cos \theta \sin \theta - 3)$
$\Rightarrow (\cos \theta + \sin \theta )(1 - 4\cos \theta \sin \theta )$
We can write $4\cos \theta \sin \theta$ as $2 \times 2\cos \theta \sin \theta$. So,
$\Rightarrow (\cos \theta + \sin \theta )(1 - 2 \times 2\cos \theta \sin \theta )$
We know that $2\cos \theta \sin \theta = \sin 2\theta$. So,
$\Rightarrow (\cos \theta + \sin \theta )(1 - 2\sin 2\theta )$
Therefore, the left side of the equation is equal to the right side of the equation.
Hence proved

(b) We have $\tan A = \dfrac{5}{6}$ and $\tan B = \dfrac{1}{{11}}$ and we have to show that $A + B = \dfrac{\pi }{4}$ or $\dfrac{{5\pi }}{4}$
Using trigonometric identity $\tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$
We have,
$\tan (A + B) = \dfrac{{\dfrac{5}{6} + \dfrac{1}{{11}}}}{{1 - \dfrac{5}{6} \times \dfrac{1}{{11}}}}$
Simplifying the above equation. We get,
$\Rightarrow \tan (A + B) = \dfrac{{\dfrac{{55 + 6}}{{66}}}}{{1 - \dfrac{5}{{66}}}}$
On further solving we get,
$\Rightarrow \tan (A + B) = \dfrac{{\dfrac{{55 + 6}}{{66}}}}{{\dfrac{{66 - 5}}{{66}}}}$
$\Rightarrow \tan (A + B) = \dfrac{{\dfrac{{61}}{{66}}}}{{\dfrac{{61}}{{66}}}}$

On dividing we get,
$\Rightarrow \tan (A + B) = \dfrac{{61}}{{66}} \times \dfrac{{66}}{{61}}$
$\Rightarrow \tan (A + B) = 1$
Shifting $\tan$ to the right side of the equation. We get,
$\Rightarrow (A + B) = {\tan ^{ - 1}}1$
We know that ${\tan ^{ - 1}}1 = \dfrac{\pi }{4}$.
$\Rightarrow (A + B) = \dfrac{\pi }{4}$
Therefore, $(A + B) = \dfrac{\pi }{4}$
Hence, proved.

Note: In order to solve these types of questions in which we have to equal both sides of the equation, first check by solving which side of the equation we can get our desired result. One should remember all trigonometric formulas before solving these types of problems. Note that some students are confused in algebraic identity ${(a + b)^3} = (a + b)({a^2} - ab + {b^2})$ there is a subtraction sign also in this identity.