If sec a=cos a + sin a
how do we prove that cos 2a= [tan(pi/2 - a) ]^2 ?
if
1/cos a = cos a + sin a
1 = oos^2 a + sin a oos a
cos^2 a = 1 - sin a cos a Needed for last step
if
sin (pi/2 -a ) = cos a
cos (pi/2- a) = sin a
so
tan(pi/2 - a) = cos a/sin a
so
[tan(pi/2 - a) ]^2 = cos^2 a/sin^2 a
cos 2a = cos^2 a - sin^2 a identity
so
prove
cos^2a/sin^2a = 1/sin^2 a - cos a/sin a
cos^2a = 1 - sin a cos a ---true
Thanks a lot!!
To prove that cos 2a = [tan(π/2 - a)]^2, we can use various trigonometric identities and properties. Here's how we can go about proving it step by step:
Step 1: Start with the given equation, sec a = cos a + sin a.
Step 2: Rewrite sec a in terms of sine and cosine using the reciprocal identity; sec a = 1/cos a.
Step 3: Substitute 1/cos a for sec a in the equation; 1/cos a = cos a + sin a.
Step 4: Multiply both sides of the equation by cos a to eliminate the denominator; 1 = cos^2 a + sin a cos a.
Step 5: Rearrange the terms; 0 = cos^2 a - 1 + sin a cos a.
Step 6: Utilize the Pythagorean identity; sin^2 a + cos^2 a = 1.
Step 7: Substitute sin^2 a = 1 - cos^2 a into the equation from the previous step; 0 = cos^2 a - 1 + cos a(√(1 - cos^2 a)).
Step 8: Simplify the equation further; 0 = cos^2 a - 1 + cos a(√(1 - cos^2 a)).
Step 9: Factor out (√(1 - cos^2 a)) from the equation; 0 = (cos a - 1)(cos a + (√(1 - cos^2 a))).
Step 10: Now, there are two possibilities for the equation to hold true:
a) cos a - 1 = 0, which implies cos a = 1.
b) cos a + (√(1 - cos^2 a)) = 0.
Step 11: For case b, simplify the equation further and solve for cos a; cos a + (√(1 - cos^2 a)) = 0
(√(1 - cos^2 a)) = -cos a
Square both sides:
1 - cos^2 a = cos^2 a
1 = 2 cos^2 a
cos^2 a = 1/2
cos a = ±√(1/2)
cos a = ±1/√2
cos a = ±(1/√2)(√2/√2)
cos a = ±(√2/2)
Step 12: Substitute the value of cos a found in step 11 into the original equation cos 2a = [tan(π/2 - a)]^2.
For cos a = (√2/2), we have a = π/4.
Therefore, cos 2a = cos (2(π/4)) = cos (π/2) = 0
And [tan(π/2 - a)]^2 = [tan(π/2 - π/4)]^2 = [tan(π/4)]^2 = 1^2 = 1
Therefore, it is true that cos 2a = [tan(π/2 - a)]^2.