Hard Trigonometry Problem with Solution for the SAT

Yesterday I gave you a Level 5 trigonometry problem for the revised SAT to try. Today I will provide a solution for this problem. If you have not yet attempted the problem go back and take a look at it first so you can try it on your own. Here is the link: Hard Trigonometry Problem for the Revised SAT

Level 5 – Trigonometry

Here is the problem once again followed by a solution:

It is given that cos x = k, where π < x < 3π/2 and x is the radian measure of an angle. If cos z = –k, which of the following could not be the value of z ?

(A) x – π
(B) π – x
(C) 2π – x
(D) 3π – x

Solution using coterminal angles and a negative identity:

cos(2π – x) = cos(x – 2π) = cos x = k ≠ –k.

So 2π – x could not be the value of z, choice (C).

Notes: (1) For the first equality we used the negative identity

cos(–A) = cos A

(see problem 104 in New SAT Math Problems), together with the fact that

x – 2π = –(2π – x) .

(2) In general we have a – b = -(b – a) . To see this simply distribute:

-(b – a) = –b + a = a – b.

(3) Using notes (1) and (2) together, we have

cos(2π – x) = cos(-(x – 2π)) = cos(x – 2π).

(4) Several more notes giving more insight into this solution can be found in New SAT Math Problems arranged by Topic and Difficulty Level.

For additional solutions to this problem such as using the cosine difference identity and using the unit circle check out New SAT Math Problems arranged by Topic and Difficulty Level.

Feel free to add your own solutions to the comments.

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