In triangle ABC, f(C)=0 and 2sin²B=cosB+cos(A−C), with f from part (a). Find sin A.

01 / Standard#Your turn

Official paper · jm01-2023 · II.4(b) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Use 0<C<π to choose the allowed branch and angle.

Hint 2

When C=π/2, A and B are complementary.

Worked solution

For w<0 the sine argument is in (−5π/6,−π/6), so f(C)<0. Hence w=1/3.
Solve the positive-branch equation in its allowed argument range.
$\sin(2C/3-\pi/6)=1/2,\quad-\pi/6<2C/3-\pi/6<\pi/2\implies C=\pi/2$
Reduce to a quadratic in sin A and keep its positive root.
$\sin^2B=\sin A\implies1-\sin^2A=\sin A\implies\sin A=\frac{\sqrt5-1}2$

sin A=(√5−1)/2.

Checks and common pitfalls: The square in sin²B denotes a power, not sin(2B).

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
For w<0 the sine argument is in (−5π/6,−π/6), so f(C)<0. Hence w=1/3.
Solve the positive-branch equation in its allowed argument range.
$\sin(2C/3-\pi/6)=1/2,\quad-\pi/6<2C/3-\pi/6<\pi/2\implies C=\pi/2$
Reduce to a quadratic in sin A and keep its positive root.
$\sin^2B=\sin A\implies1-\sin^2A=\sin A\implies\sin A=\frac{\sqrt5-1}2$

Think first. Reveal a hint when the class is ready.

2023 JM01