
Xavier J. answered 06/28/13
Tutor in Math, topics range from Algebra to Calculus.
Keeping in that csc(t) = 1/sin(t), distribute that csc(t) and you get csc(t)(sin(t) + cos(t)) = 1 + cot(t)
Angela W.
asked 06/28/13Complete the identity: csc(t)(sin(t) + cos(t)) = ?
Xavier J. answered 06/28/13
Tutor in Math, topics range from Algebra to Calculus.
Keeping in that csc(t) = 1/sin(t), distribute that csc(t) and you get csc(t)(sin(t) + cos(t)) = 1 + cot(t)
Yitzhak S. answered 07/01/13
Johns Hopkins Engineer, Specializing in Math and Science.
Whenever you have a complicated problem like this, it is best to break it down into Sines and Cosines- so
csc(t)(sin(t)+cos(t)) since csc(t) = 1/sin(t), we have
(sin(t)+cos(t))*1/(sin(t) Distributing out, we get
sin(t)/sin(t) + cos(t)/sin(t) This simplifies to
1 + cot(t)
George C. answered 06/28/13
Humboldt State and Georgetown graduate
You have a typo here.
Bam K. answered 07/01/13
Multi-discipline, Passionate, and Compassionate Tutor
Recall #1 cst t =1/sin t , cst t is the inverse of sin t
#2 cos t = cos(t/2 + t/2)=cos t/2 cos t/2 - sin t/2 sin t/2 using the formula for cos(a+b)
where a=b=t/2
#3 1 = cos t/2 cos t/2 + sin t/2 sin t/2 , The fundamental relationship in Trigonometry
rewritten with minor acceptable changes.
Therefore, csc t sin t + cos t = [1/(sin t/2)] sin t/2 + [cos t/2 cos t/2 - sin t/2 sin t/2]
= 1 + cos t/2 cos t/2 - sin t/2 sin t/2
= cos t/2 cos t/2 + sin t/2 sin t/2 + cos t/2 cos t/2 - sin t/2 sin t/2
= 2 [ cos t/2 cos t/2 ]
= 2 [ cos t/2 ]^2 , read 2 multiplied by (cos t/2) squarred!
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