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December 18, 2014

December 18, 2014

Posted by **anonymous** on Sunday, April 14, 2013 at 3:27pm.

a. cos(x)sec(x)-sin^2(x)=cos^2(x)

b. tan(x+(pi/4))= (tan(x)+1)/(1-tan(x))

c. (cos(x+y))/(cos(x-y))= (1-tan(x)tan(y))/(1+tan(x)tan(y))

d. (tan(x)+sin(x))/(1+cos(x))=tan(x)

e. (sin(x-y))/(sin(x)cos(y))=1-cot(x)tan(y)

f. sin(x)sin(y)=(1/2)[cos(x-y)-cos(x+y)]

- precalculus -
**Damon**, Sunday, April 14, 2013 at 3:44pma

1 - s^2 = c^2

yes

s^2+c^2 = 1

b

t(x+pi/4) = (t x + 1)/(1-t x)

yes

c

t x t y = s x /c x * s y/c y

so on the right we have

[1 - (s x s y / c x c y) ] / [ 1 + ( s x s y /c x c y) ]

[c x c y - s x s y] / [c x c y + s x s y]

which is indeed

c(x+y)/c(x-y)

so yes

d I am getting bored. I think you cn see the plan.

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