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If sin(pi/2-x) = -4/5 and cos(pi/2-x) = 7/8, what is the exact value of tan x?
 Mar 17, 2014
 #1
avatar+114485 
0
yeye:

If sin(pi/2-x) = -4/5 and cos(pi/2-x) = 7/8, what is the exact value of tan x?



Hi yeye,
I think that I have it but I certainly invite other people to check my logic and working.

sine and cosine are complementary ratios. That is what the co stands for in fronto of the cosine.
Now complemantary angels add up to 90 degrees or pi/2 radians,
so
sin(pi/2-x) = cos x
cos(pi/2-x) = sin x
so
cos x = -4/5 = -32/40 It is negative so x must be in the 2nd or 3rd quadrant
sin x = 7/8 = 35/40 It is positive so x must be in the 1st or 2nd quadrant
Hence x must be in the 2nd quadrant.

140317 trig triangle.JPG
From the triangle it can be seen that tan(x)=35/-32

tan(x) = - 35/32

I think that is all okay.
 Mar 17, 2014
 #2
avatar+121056 
+9
Mmmmm....we have a dilemma here. If I take the cos inverse of (-32/40), I get about 143 degrees. But, if I take the sin inverse of ( 35/40), I get about 61 degrees.- or an angle of about 119 degrees, since the angle is in the second quadrant.

The other problem is that the hypotenuse = 40. But, SQRT((-32)^2 + (35)^2) = about 47.72. So, one hypotenuse is greater than the other.

Check your problem, ara, and see if your numbers are correct. (The sin and cos inverses of your original problem produce different angles, too)

As Melody said before, someone else is free to check any flaws in my logic!!
 Mar 17, 2014
 #3
avatar+114485 
0
Thanks Chris,

Yes I did assume that the initial number were correct. I should have checked.
I still think that my method was okay.

Melody.
 Mar 18, 2014

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