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No need to apologize, you’re here to learn!

I will break this down to make it easier to follow.

Theres a few important acronym here that will help, SOHCAHTOA.  

Sin Opposite over Hypotenuse

Cosine Adjacent over Hypotenuse

Tan Opposite over Adjacent

This trick helps to remember how the trig functions are defined from a right triangle. If you want the tangent of angle B then you know its the length of the opposite side over the adjacent side!

So imagine a triangle with 3 sides: a, b, c and c is the hypotenuse

Sin(X) = a/c

Cos(X) = b/c

Tan(x) = a/b

Now it happens that if i divide sin(x) by cos(x) that is (a/c)/(b/c) = (a/c)*(c/b) = a/b = tan(x). So we can use this to evaluate the expressions!
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Draw a right triangle with acute angle 22.

Label the legs x and 1 so the side with length x is opposite the 22° angle.

So in this triangle, tan (22) = opp/adj = x/1 = x as required

Now use Pythag thm to find an expression for length of hypotenuse in this triangle. It will involve x in it.

Now write the other two triangle ratios using your side labels. This will be fractions that involve the letter x.

The sin(22) one is easiest.

The cos(158) is going to be -cos(22), because 158° + 22° = 180°, and two supplementary angles have cosine ratios that are opposites.
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Make a right triangle. Label one acute angle 22°. Label the opposite side x and adjacent side 1 because tan=OPP/ADJ.

By Pythagorean Theorem HYP=sqrt(x^2 +1)

Now use SOHCAHTOA and the fact that cos(180-22)=-cos(22).
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Hi there. just wanted to updat the post. I for some reason did not understand what you guys meant but read through my maths handbook some more later that night because i was prepping for a test and then i just understood it somehoe. has maked Trig alot easier for me and I think I did quite good on the test I wrote the next day. So just wanted to say to all you guys who commented. I really apreciate it

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