SOLUTION: (a) Express 8 Sin theta - 15 Cos theta, in the format: R.Sin(theta + beta) (b) Using your answer to part (a), solve; 8 Sin theta - 15 Cos theta=8.5 for 0degrees<=theta<=360degrees

Question 66301: (a) Express 8 Sin theta - 15 Cos theta, in the format: R.Sin(theta + beta)
(b) Using your answer to part (a), solve; 8 Sin theta - 15 Cos theta=8.5 for 0degrees<=theta<=360degrees
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
(a) Express 8 Sin theta - 15 Cos theta, in the format: R.Sin(theta + beta)
LET US USE T FOR THETA AND B FOR BETA
DIVIDE AND MULTIPLY WITH SQRT(8^2+15^2)=17
PUT 8/17=COS(B)....THEN.....15/17=SIN(B)...SINCE SIN(B)=SQRT[1-COS^2(B)
=SQRT[1-(8/17)^2]=15/17
HENCE WE GET
8SIN(T)-15COS(B)=17[(8/17)SIN(T)-(15/17)COS(T)]
=17[SIN(T)COS(B)-COS(T)SIN(B)]=17SIN[T-B]..
HENCE R=17 AND B=ARCSIN(15/17)
IF YOU WANT ONLY SIN (T+B)THEN YOU HAVE TO CHANGE THE QUADRANT
FOR THE ANGLE FROM I Q AT PRESENT.

(b) Using your answer to part (a), solve; 8 Sin theta - 15 Cos theta=8.5 for 0degrees<=theta<=360degrees
WE HAVE
17SIN(T-B)=8.5
SIN(T-B)=8.5/17=1/2=SIN(30)....OR....SIN(150).....OR...SIN(390)...ETC
T-B=30....OR....150...OR....390...0R....510...ETC
BUT B = ARCSIN(15/17)=62 DEG.
T=92...OR.....212...IN THE DESIRED RANGE OF 0 T=
T
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