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

Write 3cos⁡x–4sin⁡x3\cos x – 4\sin x in the form of k (sin⁡x−β)k\ (\sin x - \beta) .

Explanation

Solution

In this question, we need to write the given expression in the form of k (sin⁡x−β)k\ (\sin x - \beta) . First , by using the trigonometry formula sin⁡(a–b)=(sin⁡ a cos⁡ b)–(cos⁡ a sin⁡ b)\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⁡2θ+ sin⁡2θ=1 \cos^{2}\theta + \ \sin^{2}\theta = 1\ , we can find the value of kk. Then by using trigonometric ratio, tan⁡ θ=(sin⁡ θ)cos⁡ θ\tan\ \theta = \dfrac{\left( \sin\ \theta \right)}{\cos\ \theta}, we can find the value of the angle β\beta . Then we need to substitute the values of kk and β\beta , in k (sin⁡x−β)k\ (\sin x - \beta) . Using this we can write the given expression in the form of k (sin⁡x−β)k\ (\sin x - \beta) .

Complete step by step answer:
Given, 3cos⁡x–4sin⁡x3\cos x – 4\sin x
Here we need to write the expression in the form of k (sin⁡x−β)k\ (\sin x - \beta)
Thus 3cos⁡x–4sin⁡x=k (sin⁡x−β)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)\sin(a – b) = (\sin\ a\ \cos\ b - \cos\ a\ \sin\ b)
By using this trigonometry formula,
We get
⇒ 3cos⁡x–4sin⁡x=k (sin⁡ x cos⁡ β−cos⁡ a sin⁡ β)\Rightarrow \ 3\cos x – 4\sin x = k\ (\sin\ x\ \cos\ \beta - \cos\ a\ \sin\ \beta)
On multiplying kk inside,
We get,
⇒ 3cos⁡x–4sin⁡x=(k sin⁡ x cos⁡ β–k cos⁡ x sin⁡ β)\Rightarrow \ 3\cos x – 4\sin x = (k\ \sin\ x\ \cos\ \beta – k\ \cos\ x\ \sin\ \beta)
On rearranging the terms,
We get,
⇒ −4sin⁡x+3cos⁡x=(k sin⁡ x cos⁡ β–k cos⁡ x sin⁡ β)\Rightarrow \ - 4\sin x + 3\cos x = (k\ \sin\ x\ \cos\ \beta – k\ \cos\ x\ \sin\ \beta)
On equating the terms,
We get,
⇒ −4sin⁡x=ksin⁡ x cos⁡ β\Rightarrow \ - 4\sin x = k \sin\ x\ \cos\ \beta and −3cos⁡x=−k cos x sin β- 3\cos x = - k\ cos\ x\ sin\ \beta
On simplifying,
We get,
⇒ cos⁡ β=−4k\Rightarrow \ \cos\ \beta = - \dfrac{4}{k} and sin⁡ β=−3k\sin\ \beta = - \dfrac{3}{k}
We know that cos⁡2θ+ sin⁡2θ=1\cos^{2}\theta + \ \sin^{2}\theta = 1 ,
From this identity,
cos⁡2β+sin⁡2β=1\cos^{2}\beta + \sin^{2}\beta = 1
Now on substituting the values,
We get,
⇒ (−4k)2+ (−3k)2=1\Rightarrow \ \left( - \dfrac{4}{k} \right)^{2} + \ \left( - \dfrac{3}{k} \right)^{2} = 1
On simplifying,
We get,
⇒16k2+9k2=1\Rightarrow \dfrac{16}{k^{2}} + \dfrac{9}{k^{2}} = 1
By adding,
We get,
⇒25k2=1\Rightarrow \dfrac{25}{k^{2}} = 1
Thus,
⇒ k2=25\Rightarrow \ k^{2} = 25
On taking square root on both sides,
We get,
⇒ k=25\Rightarrow \ k = \sqrt{25}
On simplifying,
We get,
⇒ k=±5\Rightarrow \ k = \pm 5
Now we have found the value of kk as ±5\pm 5 .
Case 1).
Let us consider k=5k = 5
On substituting the value of kk in cos⁡ β=−4k\cos\ \beta = - \dfrac{4}{k} and sin⁡ β=−3k\sin\ \beta = - \dfrac{3}{k}
We get,
cos⁡ β=−45\cos\ \beta = - \dfrac{4}{5} and sin⁡ β=−35\sin\ \beta = - \dfrac{3}{5}
We know that
tan⁡ θ=(sin⁡ θ)cos⁡ θ\tan\ \theta = \dfrac{\left( \sin\ \theta \right)}{\cos\ \theta}
Now ,
tan⁡ β=(sin⁡ β)cos⁡ β\tan\ \beta = \dfrac{\left( {\sin\ \beta} \right)}{\cos\ \beta}
By substituting the known values,
We get,
⇒ tan⁡ β=−35−45\Rightarrow \ \tan\ \beta = \dfrac{- \dfrac{3}{5}}{- \dfrac{4}{5}}
On simplifying,
We get,
⇒ tan⁡ β=34\Rightarrow \ \tan\ \beta = \dfrac{3}{4}
On taking inverse of tan on both sides,
We get,
⇒ β=tan⁡−1(34)\Rightarrow \ \beta = \tan^{- 1}(\dfrac{3}{4})
On simplifying,
We get,
β =36.86o\beta\ = 36.86^{o}
Now we can substitute the value of kk and β\beta in equation (1) ,
We get,
3cos⁡x–4sin⁡x=5 (sin⁡x–36.9o)3\cos x – 4\sin x = 5\ (\sin x – 36.9^{o})
Case 2).
Let us consider k=−5k = - 5
On substituting the value of kk in cos⁡ β=−4k\cos\ \beta = - \dfrac{4}{k} and sin⁡ β=−3k\sin\ \beta = - \dfrac{3}{k}
We get,
cos⁡ β=45\cos\ \beta = \dfrac{4}{5} and sin⁡ β=35\sin\ \beta = \dfrac{3}{5}
Now ,
tan⁡ β=(sin⁡ β)cos⁡ β\tan\ \beta = \dfrac{\left( {\sin\ \beta} \right)}{\cos\ \beta}
By substituting the known values,
We get,
⇒ tan⁡ β=3545\Rightarrow \ \tan\ \beta = \dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}
On simplifying,
We get,
⇒ tan⁡ β=34\Rightarrow \ \tan\ \beta = \dfrac{3}{4}
On taking inverse of tan on both sides,
We get,
⇒ β=tan⁡−1(34)\Rightarrow \ \beta = \tan^{- 1}\left( \dfrac{3}{4} \right)
On simplifying,
We get,
β=36.86o\beta = 36.86^{o}
Now we can substitute the value of kk and β\beta in equation (1) ,
We get,
3cos⁡x–4sin⁡x=−5 (sin⁡x–36.9o)3\cos x – 4\sin x = - 5\ (\sin x – 36.9^{o})
Thus we can get 3cos⁡x–4sin⁡x3\cos x – 4\sin x is ±5 (sin⁡x–36.9o)\pm 5\ (\sin x – 36.9^{o}) which is in the form of k (sin⁡x−β)k\ (\sin x - \beta) .
3cos⁡x–4sin⁡x3\cos x – 4\sin x is ±5 (sin⁡x–36.9o)\pm 5\ (\sin x – 36.9^{o}) which is in the form of k (sin⁡x−β)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−β)k\ (\sin x - \beta) .