Question

express 4 cos x + 3 sin x in the form k sin ( x + alpha ) where alpha is acute hi​

Answer 1

Answer:

Assume, a △ABC△ABC right angled at B, opposite angle ∠CAB∠CAB as aa, and hypotenuse as AC. Now, presume BC is 4 units and BA is 3. So, AC becomes 5 units.

That implies, cosa=45cos⁡a=45, sina=35sin⁡a=35

Coming back to your original question.

4cosx+3sinx=k4cos⁡x+3sin⁡x=k

Divide by 5 both sides,

45cosx+35sinx=k545cos⁡x+35sin⁡x=k5

which is the form of

cosa×cosx+sina×sinx=k5cos⁡a×cos⁡x+sin⁡a×sin⁡x=k5

⟹cos(a−x)=cos(x−a)=k5⟹cos⁡(a−x)=cos⁡(x−a)=k5

Or,

k=5cos(x−a)=5cos(x+(−a))k=5cos⁡(x−a)=5cos⁡(x+(−a))

aa is approximately