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). \[ - \dfrac{1}{2}\]

Answer 1

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.

Complete step-by-step 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.