Good morning Atika! I am looking at Q6 as a supplement.

Q6 should be correct.

130 has a basic reference angle of 50 if you have drawn the diagram like I did for Q5.

Sine in both the first and second quadrants have positive output values. And 130 and 50 are main-reference angle pairs. So, sin 130 has exactly the same value as sin 50.

For sin 50, we draw a right-angled triangle in the first quadrant (the diagonal starting from the origin and pointing diagonally upwards and rightwards). We know that sin 50 = opp/hyp = k/1, so the length opposite angle 50 is k, while the hypotenuse length is 1. The other adjacent length to the triangle, using the Pythagoras' Theorem, must be sqrt (1 - k^2).

So, cos 50
= adj/hyp
= sqrt (1 - k^2) divided by 1
= sqrt (1 - k^2)

But then, cosine in the second quadrant has a negative output value, while cosine in the first quadrant has a positive output value. And 130 and 50 are main-reference angle pairs. So, cos 130 and cos 50 have opposite signs, but otherwise they are numerically equal.

So, cos 130
= - cos 50
= - sqrt (1 - k^2)