Trig expression help pls
\[\frac{ \sin^3\theta +\cos^3\theta}{ \sin \theta +\cos \theta }\]
I know sin2 + cos2 = 1
The answer in the book is \[1 - \sin \theta \cos \theta\]
Use the sum of cubes formula to factor. Hint: \(\sin \theta + \cos \theta\) is a factor.
oh ok will try
Good luck.
im stuck after getting \[(\sin \theta - \cos \theta)(\sin/^2 \theta + \sin \theta \cos \theta +\cos^2 \theta)\]
\(\sin\theta - \cos \theta\) is not a factor.
ops messed up on the writing but its sin^2 theta in the second group
did i use the sum of cubs wrong?
Sum of Cubes Formula \(x^3 + y^3 = (x + y)(x^2 - xy + y^2)\)
oh so it wud be \[(\sin \theta + \cos \theta)(\sin^2+\sin \theta \cos \theta +\cos^2 \theta)\]
\((\sin \theta + \cos \theta)(\sin^2-\sin \theta \cos \theta +\cos^2 \theta)\)
assuming that's right, then i wud use this identity? \[\sin^2 \theta + \cos^2 \theta = 1\]
oh thats where i get mixed up i guess
trying to attempt it again atm
Yes, you can use that identity to simplify
And don't forget to cancel any factors of one
so i'm confused on what to do next tbh
Do you see any factor in the numerator that matches the expression in the denominator?
ohhh yea the \[\sin \theta + \cos \theta\]
So you should have the correct result now.
yea ty so much
yw
i was spending so long on it lol ty, have a good day :D
Yep, almost everyone struggles with it at first. Then one day, it just clicks.
What's funny about these is, tan, cot, sec, and csc can always be changed to some expression with sin and cos. Meanwhile, sin and cos can always be changed to x and y. So trig is nothing but glorified algebra.
Even the most complicated expressions can be figured out.
You should write that down and commit it to memory. You never know when you'll need to simplify a very complicated trig expression on a test.
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