Solveeit Logo
Login

Question

Question: How to write \(2\sin \theta -4\cos \theta \) in the form \(r\sin \left( \theta -\alpha \right)\) ? ...

How to write 2sin⁡θ−4cos⁡θ2\sin \theta -4\cos \theta in the form rsin⁡(θ−α)r\sin \left( \theta -\alpha \right) ?
(a) Using linear formulas
(b) Using trigonometric identities
(c) Using algebraic properties
(d) None of these

Explanation

Solution

To start with, in this problem we are to write 2sin⁡θ−4cos⁡θ2\sin \theta -4\cos \theta in the form of rsin⁡(θ−α)r\sin \left( \theta -\alpha \right). We can do it by expanding the value sin⁡(θ−α)\sin \left( \theta -\alpha \right) and compare it with the given values. Then simplifying the values we will get our needed solution.

Complete step by step solution:
According to the question, we are to write 2sin⁡θ−4cos⁡θ2\sin \theta -4\cos \theta in the given form of rsin⁡(θ−α)r\sin \left( \theta -\alpha \right).
We will start with explaining the given formula, rsin⁡(θ−α)r\sin \left( \theta -\alpha \right)as rsin⁡θcos⁡α−rcos⁡θsin⁡αr\sin \theta \cos \alpha -r\cos \theta \sin \alpha .
Now, we can try to equate both equations to find the solution.
So, we can compare the coefficients of sin⁡θ\sin \theta and cos⁡θ\cos \theta and write them altogether.
Then, we get, rcos⁡α=2r\cos \alpha =2 and rsin⁡α=4r\sin \alpha =4 .
Now, to find the value r, we will square both of them and add them together.
So, we get, r2cos⁡2α=4{{r}^{2}}{{\cos }^{2}}\alpha =4and r2sin⁡2α=16{{r}^{2}}{{\sin }^{2}}\alpha =16
Thus, adding them, r2sin⁡2α+r2cos⁡2α=4+16=20{{r}^{2}}{{\sin }^{2}}\alpha +{{r}^{2}}{{\cos }^{2}}\alpha =4+16=20
From the trigonometric identities, sin⁡2α+cos⁡2α=1{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1.
Taking r2{{r}^{2}}common from the left hand side, now we have, r2(sin⁡2α+cos⁡2α)=20{{r}^{2}}\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \right)=20
Using the identity, we have, r2=20{{r}^{2}}=20
And from this we have, r=20=25r=\sqrt{20}=2\sqrt{5}
Again, dividing rsin⁡αr\sin \alpha by rcos⁡αr\cos \alpha now, we get, rsin⁡αrcos⁡α=tan⁡α\dfrac{r\sin \alpha }{r\cos \alpha }=\tan \alpha which is having a value 2.
So, we have the value of α\alpha as, α=tan⁡−12\alpha ={{\tan }^{-1}}2(using the trigonometric inverse identities).
Thus, we can put the values in rsin⁡(θ−α)r\sin \left( \theta -\alpha \right)to get our solution.
Putting back the values, we can write 2sin⁡θ−4cos⁡θ2\sin \theta -4\cos \theta as 25sin⁡(θ−tan⁡−12)2\sqrt{5}\sin \left( \theta -{{\tan }^{-1}}2 \right) .

So, the correct answer is “Option (b)”.

Note: In this problem, we have used the trigonometric angle identities to simplify and get the solution. In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables for which both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles.