21M.1.SL.TZ1.6

Note: Do not award the final A1 for proofs which work from both sides to find a common expression other than $2 \sin x \cos x-2 {\sin }^2 x$.

 

METHOD 1 (LHS to RHS)

attempt to use double angle formula for $\sin 2x$ or $\cos 2x$          M1

LHS $=2 \sin x \cos x+\cos 2x-1$  OR

$\sin 2x+1-2 {\sin }^2 x-1$  OR

$2 \sin x \cos x+1-2 {\sin }^2 x-1$

$=2 \sin x \cos x-2 {\sin }^2 x$          A1

$\sin 2x+\cos 2x-1=2 \sin x\left(\cos x-\sin x\right)=$ RHS          AG

 

METHOD 2 (RHS to LHS)

RHS $=2 \sin x \cos x-2 {\sin }^2 x$

       attempt to use double angle formula for $\sin 2x$ or $\cos 2x$          M1

       $=\sin 2x+1-2 {\sin }^2 x-1$          A1

       $=\sin 2x+\cos 2x-1=$ LHS          AG

 

[2 marks]

attempt to factorise          M1

$\left(\cos x-\sin x\right)\left(2 \sin x+1\right)=0$          A1

recognition of $\cos x=\sin x\Rightarrow \frac{\sin x}{\cos x}=\tan x=1$  OR  $\sin x=-\frac{1}{2}$        (M1)

one correct reference angle seen anywhere, accept degrees        (A1)

$\frac{\mathrm{\pi }}{4}$  OR  $\frac{\mathrm{\pi }}{6}$ (accept $-\frac{\mathrm{\pi }}{6}, \frac{7\mathrm{\pi }}{6}$)

 

Note: This (M1)(A1) is independent of the previous M1A1.

 

$x=\frac{7\mathrm{\pi }}{6},\frac{11\mathrm{\pi }}{6}, \frac{\mathrm{\pi }}{4}, \frac{5\mathrm{\pi }}{4}$          A2

 

Note: Award A1 for any two correct (radian) answers.
Award A1A0 if additional values given with the four correct (radian) answers.
Award A1A0 for four correct answers given in degrees.

 

[6 marks]

[N/A]

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