# How do you find the exact trigonometric ratios for the angle 3pi/4?

I took precalculus 2 years ago and i don't remember where to start with this type of problem. Can someone please show me what i need to do to find the lengths of the triangle and then calculate the ratios to answer the problem? Please make sure your answer is thorough as i am a bit rusty with this stuff.

• With every angle there is associated a reference angle, which is the positive acute angle that the terminal side of the given angle makes with the (positive or negative) x-axis. An angle and its reference angle have the same value for their trig functions, except possibly for sign.

3π/4 is in the second quadrant, so its cosine is negative and its sine is positive. From these you can determine the sign of the other four trig functions.

The reference angle for 3π/4 is π/4. This is an angle for which you are expected to know at least the sine and cosine: sin(π/4) = cos(π/4) = √2 / 2. Therefore,

sin(3π/4) = √2 / 2

cos(3π/4) = -√2 / 2

tan(3π/4) = sin(3π/4)/cos(3π/4) = -1

csc(3π/4) = 1/sin(3π/4) = √2

sec(3π/4) = 1/cos(3π/4) = -√2

cot(3π/4) = 1/tan(3π/4) = -1

• How To Find Trigonometric Ratios

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