Question

If sin2A = sin(90-3A). Find the value of sinA

• A.
• B.
• C.
• D. 1

Answer 1

Answer: (A), (B), (D)

The correct options are
A

B

D

1

Given sin2A = sin(90-3A)

Sin2A = cos3A

2sinA cosA = $4co{s}^{3}A-3cosA$

2sinA cosA - $4co{s}^{3}A+3cosA$ = 0

cosA (2sinA - $4co{s}^{2}A+3$) = 0

cosA = 0 or 2 sinA - $4co{s}^{2}A+3$ = 0

A = ${90}^{\circ }$

sinA = $sin{90}^{\circ }$ = 1

2sinA - 4$(1-si{n}^{2}A)+3=0$

2sinA $-4+4si{n}^{2}A+3=0$

$4si{n}^{2}A+2sinA-1=0$

$sinA=\frac{-2\pm \sqrt{(2{)}^2-4(4\left)\right(-1)}}{2\times 4}$

$sinA=\frac{-2\pm \sqrt{20}}{8}=\frac{-2\pm 2\sqrt{5}}{8}$

$sinA=\frac{-1\pm \sqrt{5}}{4}$

$sinA=1,\frac{-1\pm \sqrt{5}}{4}$