Question

If sin3A+4cosA=5, prove that sinA=3/5.

Answer 1

sine(3a) +4*kosine(a)=5

that require sine(3a)=kosine(a)=1

so 3a gotta be a+-90 deg  or pi/2 radians

3a=a+90 become 2a=90, so a=45 deg

Answer 2

sin3A+4cosA=5··· Eq.1   Since -1≦sin3A≦1, -4≦4cosA≦4, Eq.1 would be true if sin3A=1 coincided with 4cosA=4.   Check Eq.1: 4cosA=4 at A=2πn (n: any integers), but sin3A=0 at A=2πn.   Therefore, any argument A doesn't exist that satisfies the Eq.1.   There must be some misprints in Eq.1.

Here, we examine Eq.1.   Assume that right side of Eq.1 (the value 5) is the maximum value of the left, and the answer (sinA=3/5) is correct, but the left side of Eq.1 is incorrect.   Since sinA=3/5, so cosA=4/5 and tanA=3/4 (3-4-5 triangle).   Define the left side of Eq.1: y=sin3A+4cosA··· Eq.2   So that, y'=3cos3A-4sinA ··· Eq.3 and y"=-(9sin3A+4cosA) ··· Eq.4

Eq.2 takes maximum points when y'=0 coincides with y"＜0.   So, Eq.3=0 ⇒ 3cos3A-4sinA=0 ⇒ sinA/cos3A=3/4 ··· Eq.5   Note that we assume tanA=3/4.   So, try to replace cos3A in Eq.5 with cosA to find some correct equations.   We have: sinA/cosA=tanA=3/4, so sinA=3/5 and cosA=4/5   To make up for the replacement, rewrite Eq.3, Eq.4 and Eq.2.  

We have: y'=3cosA-4sinA ··· Eq.6, y"=-(3sinA+4cosA) ··· Eq.7 and y=3sinA+4cosA ··· Eq.8   Plug sinA=3/5 and cosA=4/5 into Eq.6, Eq.7 and Eq.8.   We have: y'=0, y"=-5＜0 and y=5   So, Eq.8 takes its maximum value when sinA=3/5.

Therefore, 3sinA+4cosA=5 could be ONE of the correct equations, and the answer for the question is sinA=3/5.