The expression 2cos(135°) – sin(150°) + 4tg(225°) – 3cos(660°) is equal to:

Alternatives

A: 4 - √2

B: 2 - √2

C: √2 - 2

D: 2 - (√2)/2

E: 1

Hi!

I used x° * pi/180 = x rad, as well as the trigonometric circle properties

2cos(135°)-sin(150°)+4tan(225°)-3cos(660°)

= 2cos(3pi/4)-sin(5pi/6)+4tan(5pi/4)-3cos(11pi/3)

= 2cos(-5pi/4)-sin(-7pi/6)+4tan(pi/4)-3cos(2pi/3)

= 2cos(-pi/4)-sin(-pi/6)+4tan(pi/4)-3cos(-pi/3)

= 2 * -square root of 2/2 -1/2 + 4 * 1 - 3 *1/2

= 2 - square root of 2

@AriHoresh i couldn’t find the correct answer when using the formulas -sin(x)=sin(-x) and cos(-x)=cos(x), does that mean they are incorrect?

Also, is this in the math syllabus?

Thank you!

I couldn’t see similar questions in the IMAT’s syllabus, only the BMAT’s and these questions got excluded as well over the years.

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