Question

If \[\cos {\text{ }}(\alpha {\text{ + }}\beta {\text{) = 0}}\], then \[{\text{sin }}(\alpha {\text{ - }}\beta {\text{)}}\] can be reduced to
A. $\cos \beta $
B. $\cos 2\beta $
C. $\sin \alpha $
D. $\sin 2\alpha $

Answer 1

Hint: To solve this question, we will use the trigonometric value of $\cos {90^0}$ which is equal to zero. We will apply this in the given condition and from it we will find the value of $\alpha $ to find the value of \[{\text{sin }}(\alpha {\text{ - }}\beta {\text{)}}\].

Complete step-by-step answer:

Now, we are given \[\cos {\text{ }}(\alpha {\text{ + }}\beta {\text{) = 0}}\]. Now, at ${90^0}$, cos x = 0. So, we can write as

\[\cos {\text{ }}(\alpha {\text{ + }}\beta {\text{) = cos 9}}{{\text{0}}^0}\]

As, in the above equation, terms on both sides are of the form cos x. So, removing cos from the above equation, we get

$\alpha {\text{ + }}\beta {\text{ = 9}}{{\text{0}}^0}$. So, from this equation, we can find the value of $\alpha $. So,

$\alpha {\text{ = 9}}{{\text{0}}^0}{\text{ - }}\beta $

Now, we will put this value of $\alpha $ in \[{\text{sin }}(\alpha {\text{ - }}\beta {\text{)}}\], we get

\[{\text{sin }}(\alpha {\text{ - }}\beta {\text{)}}\] = \[{\text{sin }}({90^0}{\text{ - }}\beta {\text{ - }}\beta {\text{)}}\]

\[{\text{sin }}(\alpha {\text{ - }}\beta {\text{)}}\] = $\sin {\text{ (9}}{{\text{0}}^0}{\text{ - 2}}\beta {\text{)}}$

Now, we know that for x lying in the first quadrant, we have $\sin {\text{ (9}}{{\text{0}}^0}{\text{ - x) = cos x }}$.

So, we can write as,

\[{\text{sin }}(\alpha {\text{ - }}\beta {\text{)}}\] = $\sin {\text{ (9}}{{\text{0}}^0}{\text{ - 2}}\beta {\text{)}}$ = $\cos 2\beta $

Therefore, \[{\text{sin }}(\alpha {\text{ - }}\beta {\text{)}}\] = $\cos 2\beta $

So, option (B) is correct.

Note: When we come up with such types of questions, where we are given a trigonometric term on one side and a numerical value on the other side, we will always use the trigonometric values to find the solution. In this question, cos x can have a value equal to zero at ${180^0}$ so we get $\alpha {\text{ + }}\beta {\text{ = 18}}{{\text{0}}^0}$ and $\alpha {\text{ = 18}}{{\text{0}}^0}{\text{ - }}\beta $.When we put this value in \[{\text{sin }}(\alpha {\text{ - }}\beta {\text{)}}\], we will get the same answer because in the second quadrant, $\sin {\text{ (18}}{{\text{0}}^0}{\text{ - x) = cos x }}$.