Question

if a cos theta + b sin theta equal 4 and a sin theta minus B cos theta equal 3 then a square + b square = 25

Answer 1

Answer:

Step-by-step explanation:

A cos theta + B sin theta = 4

[A cos theta + B sin theta]^2 = 16 ……….i

A sin theta - B cos theta = 3

[A sin theta - B cos theta]^2 = 9 ………….i i

Adding equation i and ii we get,

(A cos theta)^2 + (B sin theta)^2 + (A sin theta)^2 + (B cos theta)^2 + 2 AB cos theta sin theta - 2 AB cos theta sin theta = 25

A^2[(cos theta)^2 + (sin theta)^2] + B^2[(cos theta)^2 + (sin theta)^2] = 25

A^2 + B^2 = 25

Answer 2

Let theta =x

a.cosx+b.sinx=4……………….(1)

a.sinx-b.cosx=3………………….(2)o

On squaring the eqn.(1) & (2) both sides and then adding.

a^2.cos^2x+b^2.sin^2x+2.a.b.sinx.cosx+a^2sin^2x+b^2.cos^2x

-2.a.b.sinx.cosx =4^2+3^2.

or. a^2(cos^2x+sin^2x)+b^2.(sin^2x+cos^2x)=25

or. a^2.1+b^2.1=25

or. a^2+b^2 = 25. Answer.