Question

Write \[3\cos x – 4\sin x\] in the form of \[k\ (\sin x - \beta)\] .

Answer 1

Hint:In this question, we need to write the given expression in the form of \[k\ (\sin x - \beta)\] . First , by using the trigonometry formula \[\sin(a – b) = (\sin\ a\ \cos\ b) – (\cos\ a\ \sin\ b) \] , we can expand the expression. Then on comparing the terms and simplifying using the identity \[\cos^{2}\theta + \ \sin^{2}\theta = 1\ \] , we can find the value of \[k\]. Then by using trigonometric ratio, \[\tan\ \theta = \dfrac{\left( \sin\ \theta \right)}{\cos\ \theta}\], we can find the value of the angle \[\beta\] . Then we need to substitute the values of \[k\] and \[\beta\] , in \[k\ (\sin x - \beta)\] . Using this we can write the given expression in the form of \[k\ (\sin x - \beta)\] .

Complete step by step answer:
Given, \[3\cos x – 4\sin x\]
Here we need to write the expression in the form of \[k\ (\sin x - \beta)\]
Thus \[3\cos x – 4\sin x = k\ (\sin x - \beta)\] ••• (1)
We know that \[\sin(a – b) = (\sin\ a\ \cos\ b - \cos\ a\ \sin\ b) \]
By using this trigonometry formula,
We get
\[\Rightarrow \ 3\cos x – 4\sin x = k\ (\sin\ x\ \cos\ \beta - \cos\ a\ \sin\ \beta)\]
On multiplying \[k\] inside,
We get,
\[\Rightarrow \ 3\cos x – 4\sin x = (k\ \sin\ x\ \cos\ \beta – k\ \cos\ x\ \sin\ \beta)\]
On rearranging the terms,
We get,
\[\Rightarrow \ - 4\sin x + 3\cos x = (k\ \sin\ x\ \cos\ \beta – k\ \cos\ x\ \sin\ \beta)\]
On equating the terms,
We get,
\[\Rightarrow \ - 4\sin x = k \sin\ x\ \cos\ \beta\] and \[- 3\cos x = - k\ cos\ x\ sin\ \beta\]
On simplifying,
We get,
\[\Rightarrow \ \cos\ \beta = - \dfrac{4}{k}\] and \[\sin\ \beta = - \dfrac{3}{k}\]
We know that \[\cos^{2}\theta + \ \sin^{2}\theta = 1\] ,
From this identity,
\[\cos^{2}\beta + \sin^{2}\beta = 1\]
Now on substituting the values,
We get,
\[\Rightarrow \ \left( - \dfrac{4}{k} \right)^{2} + \ \left( - \dfrac{3}{k} \right)^{2} = 1\]
On simplifying,
We get,
\[\Rightarrow \dfrac{16}{k^{2}} + \dfrac{9}{k^{2}} = 1\]
By adding,
We get,
\[\Rightarrow \dfrac{25}{k^{2}} = 1\]
Thus,
\[\Rightarrow \ k^{2} = 25\]
On taking square root on both sides,
We get,
\[\Rightarrow \ k = \sqrt{25}\]
On simplifying,
We get,
\[\Rightarrow \ k = \pm 5\]
Now we have found the value of \[k\] as \[\pm 5\] .
Case 1).
Let us consider \[k = 5\]
On substituting the value of \[k\] in \[\cos\ \beta = - \dfrac{4}{k}\] and \[\sin\ \beta = - \dfrac{3}{k}\]
We get,
\[\cos\ \beta = - \dfrac{4}{5}\] and \[\sin\ \beta = - \dfrac{3}{5}\]
We know that
\[\tan\ \theta = \dfrac{\left( \sin\ \theta \right)}{\cos\ \theta}\]
Now ,
\[\tan\ \beta = \dfrac{\left( {\sin\ \beta} \right)}{\cos\ \beta}\]
By substituting the known values,
We get,
\[\Rightarrow \ \tan\ \beta = \dfrac{- \dfrac{3}{5}}{- \dfrac{4}{5}}\]
On simplifying,
We get,
\[\Rightarrow \ \tan\ \beta = \dfrac{3}{4}\]
On taking inverse of tan on both sides,
We get,
\[\Rightarrow \ \beta = \tan^{- 1}(\dfrac{3}{4})\]
On simplifying,
We get,
\[\beta\ = 36.86^{o}\]
Now we can substitute the value of \[k\] and \[\beta\] in equation (1) ,
We get,
\[3\cos x – 4\sin x = 5\ (\sin x – 36.9^{o})\]
Case 2).
Let us consider \[k = - 5\]
On substituting the value of \[k\] in \[\cos\ \beta = - \dfrac{4}{k}\] and \[\sin\ \beta = - \dfrac{3}{k}\]
We get,
\[\cos\ \beta = \dfrac{4}{5}\] and \[\sin\ \beta = \dfrac{3}{5}\]
Now ,
\[\tan\ \beta = \dfrac{\left( {\sin\ \beta} \right)}{\cos\ \beta}\]
By substituting the known values,
We get,
\[\Rightarrow \ \tan\ \beta = \dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}\]
On simplifying,
We get,
\[\Rightarrow \ \tan\ \beta = \dfrac{3}{4}\]
On taking inverse of tan on both sides,
We get,
\[\Rightarrow \ \beta = \tan^{- 1}\left( \dfrac{3}{4} \right)\]
On simplifying,
We get,
\[\beta = 36.86^{o}\]
Now we can substitute the value of \[k\] and \[\beta\] in equation (1) ,
We get,
\[3\cos x – 4\sin x = - 5\ (\sin x – 36.9^{o})\]
Thus we can get \[3\cos x – 4\sin x\] is \[\pm 5\ (\sin x – 36.9^{o})\] which is in the form of \[k\ (\sin x - \beta)\] .
\[3\cos x – 4\sin x\] is \[\pm 5\ (\sin x – 36.9^{o})\] which is in the form of \[k\ (\sin x - \beta)\] .

Note:These types of questions require grip over the concepts of trigonometry and identity . In this question , We are provided with a trigonometric expression in sine and cosine, then we need to use the formula and identity which contains both the given trigonometric function. While solving such trigonometric identities problems, we need to have a good knowledge about the trigonometric identities. One must know the correct trigonometric formulas and ratios to write the given trigonometric expressions of the form \[k\ (\sin x - \beta)\] .