Question

How do you evaluate $\dfrac{{{\sin }^{2}}{{15}^{\circ }}+\sin {{75}^{\circ }}}{{{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}}$ ?

Answer 1

We can do this problem by separately evaluating the numerator and denominator. Then substitute the values and then divide both the values to get the final answer for the given equation.

Complete step by step solution:
According to the problem, we are asked to evaluate $\dfrac{{{\sin }^{2}}{{15}^{\circ }}+\sin {{75}^{\circ }}}{{{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}}$. We take the equation as equation 1.

$\dfrac{{{\sin }^{2}}{{15}^{\circ }}+\sin {{75}^{\circ }}}{{{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}}$--- ( 1 )

Now we solve the numerator and the denominator separately.
Numerator = ${{\sin }^{2}}{{15}^{\circ }}+{{\sin }^{2}}{{75}^{\circ }}$ ---- (2)

Denominator = ${{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}$ ---- (3)

First we solve the numerator:
We know that sin75 = cos15. Therefore substituting in equation 2, we get:
$\Rightarrow {{\sin }^{2}}{{15}^{\circ }}+{{\sin }^{2}}{{75}^{\circ }}={{\sin }^{2}}{{15}^{\circ }}+{{\cos }^{2}}{{15}^{\circ }}$ ----- (4)

But we know that, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Therefore, substituting it in equation 4, we get:

Here we take the value of x as 15. x = 15.
$\Rightarrow {{\sin }^{2}}{{15}^{\circ }}+\sin {{75}^{\circ }}={{\sin }^{2}}{{15}^{\circ }}+\cos {{15}^{\circ }}$

$\Rightarrow {{\sin }^{2}}{{15}^{\circ }}+\sin {{75}^{\circ }}=1$---- (6)

Now we solve the denominator:
We know that cos54 = sin36. Therefore, substituting in equation 3, we get:
$\Rightarrow {{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}={{\cos }^{2}}{{36}^{\circ }}+{{\sin }^{2}}{{36}^{\circ }}$ ----- (5)

But we know that, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Therefore, substituting it in equation 5, we get:

Here we take the value of x as 36. x = 3.
$\Rightarrow {{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}={{\cos }^{2}}{{36}^{\circ }}+{{\sin }^{2}}{{36}^{\circ }}$

$\Rightarrow {{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}=1$--- (7)

Therefore, substituting 6 and 7 in equation 1, we get
$\Rightarrow \dfrac{{{\sin }^{2}}{{15}^{\circ }}+{{\sin }^{2}}{{75}^{\circ }}}{{{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}}=\dfrac{1}{1}=1$

$\Rightarrow \dfrac{{{\sin }^{2}}{{15}^{\circ }}+{{\sin }^{2}}{{75}^{\circ }}}{{{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}}=1$ ---- Final answer

So, we have found the derivative of the given equation $\dfrac{{{\sin }^{2}}{{15}^{\circ }}+\sin {{75}^{\circ }}}{{{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}}$ as $\dfrac{{{\sin }^{2}}{{15}^{\circ }}+{{\sin }^{2}}{{75}^{\circ }}}{{{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}}=1$.

Therefore, the solution of the given equation $\dfrac{{{\sin }^{2}}{{15}^{\circ }}+\sin {{75}^{\circ }}}{{{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}}$ is $\dfrac{{{\sin }^{2}}{{15}^{\circ }}+{{\sin }^{2}}{{75}^{\circ }}}{{{\cos }^{2}}{{36}^{\circ }}+{{\cos }^{2}}{{54}^{\circ }}}=1$.

Note: In this question, we should be careful while converting sin to cos and cos to sin. You should also remember basic trigonometric formulas to solve this. IN the above question, instead of converting sin75 to cos15, we can convert sin15 to cos75. Even then, we get the same answer.