Question

Prove the the following trigonometric functions: ${\text{cos1}}{{\text{8}}^0} - {\text{ sin1}}{{\text{8}}^0} = \sqrt 2 {\text{sin2}}{{\text{7}}^0}$.

Answer 1

Hint - Divide the entire L.H.S of the equation by $\sqrt 2$ and then use trigonometric formula Sin (A – B) to convert the entire equation in terms of the trigonometric function Sine.

Complete step by step answer:
Let’s get started by proving the L.H.S is equal to R.H.S of the given equation.
L.H.S
$\Rightarrow {\text{cos1}}{{\text{8}}^0} - {\text{sin1}}{{\text{8}}^0}$
Divide the entire equation with $\sqrt 2$
$\Rightarrow \dfrac{1}{{\sqrt 2 }}{\text{cos1}}{{\text{8}}^0} - \dfrac{1}{{\sqrt 2 }}{\text{sin1}}{{\text{8}}^0}$
We know that ${\text{cos4}}{{\text{5}}^0} = {\text{sin4}}{{\text{5}}^0} = \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow {\text{sin4}}{{\text{5}}^0}{\text{cos1}}{{\text{8}}^0} - {\text{cos4}}{{\text{5}}^0}{\text{sin1}}{{\text{8}}^0}{\text{ }}$ --- Equation 1
Using the trigonometric formula,
Sin (A-B) = SinACosB – CosASinB.
⟹Here A = ${\text{4}}{{\text{5}}^0}$and B =${\text{1}}{{\text{8}}^0}$, now Equation 1 becomes
$\Rightarrow {\text{Sin}}\left( {{{45}^0} - {{18}^0}} \right) = {\text{ Sin2}}{{\text{7}}^0}$
Now we obtained, $\dfrac{1}{{\sqrt 2 }}{\text{cos1}}{{\text{8}}^0} - \dfrac{1}{{\sqrt 2 }}{\text{sin1}}{{\text{8}}^0} = {\text{Sin2}}{{\text{7}}^0}$
$\Rightarrow {\text{cos1}}{{\text{8}}^0} - {\text{sin1}}{{\text{8}}^0} = \sqrt 2 {\text{sin2}}{{\text{7}}^0}$
Which is equal to the R.H.S, hence proved.

Note – In such problems, the trick is to transform L.H.S equations, by using trigonometric formulae to convert the entire equation into desired trigonometric ratio present in the R.H.S. Basic trigonometric formulae and tables are necessary to approach the solution.