Let $\sin {100^ \circ } = m$. Find an expression for $\cos {100^ \circ }$ in terms of m.

Let $\sin {100^ \circ } = m$ . Find an expression for $\tan {100^ \circ }$ in terms of m.

Let $\sin {100^ \circ } = m$. Find an expression for $\sin {200^ \circ }$ in terms of m.

Note: All answers must be given in terms of m. If a candidate makes an error that means there is no m in their answer, do not award the final A1FT mark.

METHOD 1 

valid approach involving Pythagoras     (M1)

e.g. ${\sin ^2}x + {\cos ^2}x = 1$ , labelled diagram 

correct working (may be on diagram)     (A1)

e.g. ${m^2} + {(\cos 100)^2} = 1$ , $\sqrt {1 - {m^2}}$

$\cos 100 = - \sqrt {1 - {m^2}}$    A1     N2

[3 marks]

METHOD 2

valid approach involving tan identity     (M1)

e.g. $\tan  = \frac{{\sin }}{{\cos }}$

correct working     (A1)

e.g. $\cos 100 = \frac{{\sin 100}}{{\tan 100}}$

$\cos 100 = \frac{m}{{\tan 100}}$     A1     N2

[3 marks]

METHOD 1

$\tan 100 = - \frac{m}{{\sqrt {1 - {m^2}} }}$ (accept $\frac{m}{{ - \sqrt {1 - {m^2}} }}$)     A1     N1

[1 mark]

METHOD 2

$\tan 100 = \frac{m}{{\cos 100}}$     A1     N1

[1 mark]

METHOD 1

valid approach involving double angle formula     (M1)

e.g. $\sin 2\theta = 2\sin \theta cos\theta$

$\sin 200 = - 2m\sqrt {1 - {m^2}}$  (accept $2m\left( { - \sqrt {1 - {m^2}} } \right)$)     A1     N2

Note: If candidates find $\cos 100 = \sqrt {1 - {m^2}}$ , award full FT in parts (b) and (c), even though the values may not have appropriate signs for the angles.

[2 marks]

METHOD 2

valid approach involving double angle formula     (M1)

e.g. $\sin 2\theta  = 2\sin \theta \cos \theta$ , $2m \times \frac{m}{{\tan 100}}$

$\sin 200 = \frac{{2{m^2}}}{{\tan 100}}( = 2m\cos 100)$     A1     N2

[2 marks]

While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write $\sqrt {1 - {m^2}} = 1 - m$ . In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because $\sin {100^ \circ } = m$ , then $\sin {200^ \circ } = 2m$ . In addition, some candidates did not seem to understand what writing an expression "in terms of m" meant.

While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write $\sqrt {1 - {m^2}} = 1 - m$ . In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because $\sin {100^ \circ } = m$ , then $\sin {200^ \circ } = 2m$ . In addition, some candidates did not seem to understand what writing an expression "in terms of m" meant.

While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write $\sqrt {1 - {m^2}} = 1 - m$ . In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because $\sin {100^ \circ } = m$ , then $\sin {200^ \circ } = 2m$ . In addition, some candidates did not seem to understand what writing an expression "in terms of m" meant.