If you can pass this quiz, you are ready to take this course
- Evaluate
in terms of
![{\displaystyle \tan(\theta )=\sin(\theta )/{\sqrt {(}}1-(\sin ^{2}(\theta ))\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b627f4a08221664bcba43d369d657302f5775149)
- Shyam (T/C)
- If
then what does
equal?
where ![{\displaystyle x=[-1,1]\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cfb9a0c09b2c153088afdbe63c4b4146b0c8c862)
- Shyam (T/C)
- Prove
using
![{\displaystyle \sin ^{2}(\theta )+cos^{2}(\theta )=1\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/50a25d6127d16a438bb6ef9e14d5a82e8699b48d)
- divide both sides by
![{\displaystyle \cos ^{2}(\theta )=>\sin ^{2}(\theta )/cos^{2}(\theta )+1=1/cos^{2}(\theta )\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c22e0bf15459ce28318eeb970dc4f3b2011c680)
![{\displaystyle =>\tan ^{2}(\theta )+1=sec^{2}(\theta )\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb927f208800f2d2899aa5d985c56a124a9847d1)
- Shyam (T/C)
- Find the double angle idenities for the cosine function using the above rule.
- replace
by ![{\displaystyle A=>\cos(A+A)=\cos(A)\cos(A)-\sin(A)\sin(A)\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/4569afcbe569a2d53aca9a3bfed52eea85952c0b)
![{\displaystyle =>\cos(2A)=\cos ^{2}(A)-\sin ^{2}(A)\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/55f8a5f59afcea0c1f0b4b90272933c94a814adb)
- Shyam (T/C)
- Find the half angle idenities from the double angle idenities.
![{\displaystyle =>\cos(2A)=\cos ^{2}(A)-\sin ^{2}(A)\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/55f8a5f59afcea0c1f0b4b90272933c94a814adb)
- replace
by
using ![{\displaystyle \sin ^{2}(A)+\cos ^{2}(A)=1\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0313314fc64fb82132cb848e3d8d3855daee4c16)
- Shyam (T/C)
- Find the value of
without exponents using the above rules
![{\displaystyle =>\cos(2\theta )=2\cos ^{2}(\theta )-1\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0bcd9bf92f99cbf9bd140b68f6cb8bb852e5d8ca)
![{\displaystyle =>\cos ^{2}(\theta )=(1+\cos(2\theta ))/2\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3dd78135dede69ef700694b78297743954c8128)
- Shyam (T/C)
- (Challenge) Find the value of
without exponents
![{\displaystyle \cos(3\theta )=cos(\theta +2\theta )\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9af1e98fb42af2490ee63c0faac57a514f6f0a8)
![{\displaystyle =>\cos(3\theta )=cos(\theta )\cos(2\theta )-sin(\theta )\sin(2\theta )\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b22917eea25a37139dd798d37de0092470b6a4e7)
using
and ![{\displaystyle \sin(2\theta )=2\sin(\theta )\cos(\theta )\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/29dbe4984c5a9b0041e778a7554e8f28a495b662)
![{\displaystyle =>\cos(3\theta )=cos(\theta )((2\cos ^{2}(\theta )-1)-2sin^{2}(\theta ))\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a1ac83698bb545fae6189ff93c607be20bbfcc4)
using ![{\displaystyle \,\ \sin ^{2}(\theta )+\cos ^{2}(\theta )=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de7112977553532350dd51593203880ceaece307)
![{\displaystyle =>4\cos ^{3}(\theta )=cos(3\theta )+3cos(\theta )\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a9b6d859e7fa9898c7f286f391d0c400184e94e)
![{\displaystyle =>\cos ^{3}(\theta )=(cos(3\theta )+3cos(\theta ))/4\,\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6a504921db58c3b171245b47894eeec57199817)
- Shyam (T/C) 19:42, 18 November 2006 (UTC)