Prove the following trig identities starting with the left side.
1) sin^2x + cos^2x = 1
2) (cosx - sinx)^2 + (cosx + sinx)^2 = 2
3) cotx + tanx = secx*cscx
Thank you for any help!
Prove the following trig identities starting with the left side.
1) sin^2x + cos^2x = 1
2) (cosx - sinx)^2 + (cosx + sinx)^2 = 2
3) cotx + tanx = secx*cscx
Thank you for any help!
1. This seems very weird, it's pretty much the basic proof.
$\displaystyle sin(x) = opp/hyp$
$\displaystyle cos(x) = adj/hyp$
Therefore if we square both:
$\displaystyle sin^2(x) = (opp)^2/(hyp)^2$
$\displaystyle cos^2(x) = (adj)^2/(hyp)^2$
Adding them:
$\displaystyle sin^2(x) + cos^2(x) = \frac{(opp)^2 + (adj)^2}{(hyp)^2}$
By the definition of a right angled triangle and using Pythagoras the right side cancels to 1.
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2. Expand to give:
$\displaystyle cos^2(x) - 2sin(x)cos(x) + sin^2(x) + cos^2(x) + 2sin(x)cos(x) + cos^2(x)$
Remember what $\displaystyle cos^2(x)+sin^2(x)$ is equal to
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3. Rewrite in terms of sin and cos:
$\displaystyle \frac{cos(x)}{sin(x)} + \frac{sin(x)}{cos(x)}$
Give them the same denominator by cross multiplying
$\displaystyle \frac{cos^2(x) + sin^2(x)}{sin(x)cos(x)}$
and cancel to give the rhs