Question

If sin x + cos x = t, then sin 3x - cos 3x is equal to

• A. 3t (2$t^2$ + 1)
• B. t (2$t^2$ - 3)
• C. t (2$t^2$ + 3)
• D. -2t $(3t^2$ + 1)

Answer 1

The correct option is B

t (2$t^2$ - 3)

We have to simplify/modify sin 3x - cos 3x to express in terms of sin x + cos x.

We will express them in terms of sin x and cos x
sin 3x - cos 3x = 3 sin x - 4 $si{n}^3$x - 4 $co{s}^3$ x + 3 cos x

= 3 (sin x+ cos x) - 4 ($si{n}^3$ x + $co{s}^3$ x)

= 3(sin x + cos x) - 4 [ $(sinx+cosx{)}^3$ - 3 sin x cos x (sin x + cos x)]
We know, $t^2$ = $(sinx+cosx{)}^2$ =1+ 2 sin x cos x

⇒ sin x cos x =$\frac{t^2-1}{2}$

⇒ sin 3x - cos 3x = 3t - 4[ $t^3$ - 3 $\frac{(t^2-1)}{2}$t]

= 3t - 4[ $t^3$ - $\frac{3t^3}{2}$ + $\frac{3t}{2}$ ]

= 3t - 4[- $\frac{t^3}{2}$ + $\frac{3t}{2}$ ]

= 3t + 2$t^3$ - 6t

= 2$t^3$ - 3t

= t (2$t^2$ - 3 )