Question

${\text{sin}}\left( {{\text{13}}{{\text{5}}^{\text{o}}}{\text{ - x}}} \right){\text{ + sin(13}}{{\text{5}}^{\text{o}}}{\text{x)}}$ equals-
A. $\sqrt {\text{2}} {\text{sinx}}$
B. -1
C. $\sqrt 3 {\text{cosx}}$
D. $\sqrt 2 {\text{cosx}}$
E.$\sqrt 3 {\text{sinx}}$

Answer 1

Hint-In this question we use the basic theory or formula related to trigonometric functions. Learning these formulas will help students to score good marks in this concept. They can find the trigonometry table along with inverse trigonometry formulas to solve the problems based on them. Some of them are mentioned below-

Complete step-by-step answer:

As we know, basic formulas:

sin(x+y) = sin(x)cos(y)+cos(x)sin(y)

cos(x+y) = cos(x)cos(y)–sin(x)sin(y)

sin(${\text{9}}{{\text{0}}^{\text{o}}}$+x) = cosx

sin(90°−x) = cos x

cos(90°−x) = sin x

Now, we have

Given expression-

${\text{sin}}\left( {{\text{13}}{{\text{5}}^{\text{o}}}{\text{ - x}}} \right){\text{ + sin(13}}{{\text{5}}^{\text{o}}}{\text{x)}}$

Sin(${\text{13}}{{\text{5}}^{\text{o}}}$) cosx – cos(${\text{13}}{{\text{5}}^{\text{o}}}$) sinx +sin(${\text{13}}{{\text{5}}^{\text{o}}}$x)

=$\dfrac{{{\text{cosx}}}}{{\sqrt {\text{2}} }}{\text{ - }}\dfrac{{{\text{sinx}}}}{{\sqrt 2 }}$+$\dfrac{{{\text{cosx}}}}{{\sqrt {\text{2}} }} + \dfrac{{{\text{sinx}}}}{{\sqrt 2 }}$

=$\sqrt 2 {\text{cosx}}$

Therefore, ${\text{sin}}\left( {{\text{13}}{{\text{5}}^{\text{o}}}{\text{ - x}}} \right){\text{ + sin(13}}{{\text{5}}^{\text{o}}}{\text{x)}}$equals $\sqrt 2 {\text{cosx}}$.

So, option (D) is correct.

Note-Before solving this question you must remember the trigonometric identities that are used while solving the question. There is an enormous number of uses of trigonometry and its formulae. For example, the technique of triangulation is used in Geography to measure the distance between landmarks; in Astronomy, to measure the distance to nearby stars and also in satellite navigation systems.