QUESTION

# Solve $\dfrac{{{{\cos }^2}33^\circ - {{\cos }^2}57^\circ }}{{\sin 21^\circ - \cos 21^\circ }}$.(a). $- \sqrt 2$(b). $\dfrac{1}{2}$(c). $\dfrac{1}{{\sqrt 2 }}$(d). $- \dfrac{1}{2}$

Hint: We can convert cosine to sine by using the formula $\cos x = \sin (90^\circ - x)$. Then use the formula ${\cos ^2}x - {\sin ^2}x = \cos 2x$. Then we can write sinA – cosB in the form of cosine alone by introducing a factor of $\dfrac{1}{{\sqrt 2 }}$. Then simplify to get the answer.

Let us assign the given expression to the variable A as follows:

$A = \dfrac{{{{\cos }^2}33^\circ - {{\cos }^2}57^\circ }}{{\sin 21^\circ - \cos 21^\circ }}$
To simplify this expression, we convert the numerator and denominator purely in terms of cosine alone and then evaluate.

Now, we convert cos57° in terms of sine using the formula $\cos x = \sin (90^\circ - x)$. Then we have:

$A = \dfrac{{{{\cos }^2}33^\circ - {{\sin }^2}(90^\circ - 57^\circ )}}{{\sin 21^\circ - \cos 21^\circ }}$

Simplifying the above expression, we have:

$A = \dfrac{{{{\cos }^2}33^\circ - {{\sin }^2}33^\circ }}{{\sin 21^\circ - \cos 21^\circ }}$

We know that the value of ${\cos ^2}x - {\sin ^2}x$ is equal to $\cos 2x$. Then, we have:

$A = \dfrac{{\cos (2 \times 33^\circ )}}{{\sin 21^\circ - \cos 21^\circ }}$

Simplifying the above expression, we have:

$A = \dfrac{{\cos 66^\circ }}{{\sin 21^\circ - \cos 21^\circ }}$

Next, we multiply and divide the denominator of the expression by $\sqrt 2$. Then, we have:

$A = \dfrac{{\cos 66^\circ }}{{\dfrac{{\sqrt 2 }}{{\sqrt 2 }}(\sin 21^\circ - \cos 21^\circ )}}$

Then, we take the term $\dfrac{1}{{\sqrt 2 }}$ inside the bracket and multiply with $\sin 21^\circ$ and $\cos 21^\circ$.

$A = \dfrac{{\cos 66^\circ }}{{\sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\sin 21^\circ - \dfrac{1}{{\sqrt 2 }}\cos 21^\circ } \right)}}$

We know that the value of sin45° and cos45° is $\dfrac{1}{{\sqrt 2 }}$, then, we have:

$A = \dfrac{{\cos 66^\circ }}{{\sqrt 2 \left( {\sin 45^\circ \sin 21^\circ - \cos 45^\circ \cos 21^\circ } \right)}}$

We know that $\cos x\cos y - \sin x\sin y = \cos (x + y)$, then we have the following:

$A = \dfrac{{\cos 66^\circ }}{{\sqrt 2 \left( { - \cos (45^\circ + 21^\circ )} \right)}}$

Simplifying the above expression, we have:

$A = \dfrac{{\cos 66^\circ }}{{\sqrt 2 \left( { - \cos 66^\circ } \right)}}$

Canceling the common terms, we have:

$A = \dfrac{1}{{\sqrt 2 \left( { - 1} \right)}}$

Simplifying, we have:

$A = - \dfrac{1}{{\sqrt 2 }}$

Then, the final value of the given expression is $- \dfrac{1}{{\sqrt 2 }}$.

Hence, the correct answer is option (c).

Note: It is a tricky question and you can also use the formula ${a^2} - {b^2} = (a + b)(a - b)$ to evaluate the numerator and then simplify using the sum of cosines.