Question

Prove That
Sin^0.5(x) * cos(x) - sin^(5/2)(x) cos(x) = cos^3(x)* squareroot(sinx)

A better version in the post below

Answer 1

\[\sin ^{.5} * \cos(x) - \sin ^{5/2}x * \cos(x) = \cos ^{3} * \sqrt{sinx}\]

Answer 2

Did you try to factor sqrt(sin(x))*cos(x) out on the left hand side

Answer 3

Do you know why I'm asking you to try that?

Answer 4

That one side is written has a product and we need this side as a product. Factoring will do that. I notice they both have cos(x) and common and I know I need a sqrt(sin(x)) also because of the other side so I factored those items out.
Let me know when you have tried this.

Answer 5

How does one factor out sqrt(sinx) * cos(x) from sqrt(sin^5x) cosx....

Answer 6

1/2+?=5/2

Answer 7

\[u^\frac{1}{2} \cdot u^\frac{4}{2}=u^\frac{5}{2}\]

Answer 8

Do you understand? I'm just using the law of exponents.

Answer 9

Yes, it's confusing when one adds trig to it for some reason

Answer 10

\[\sin^\frac{1}{2}(x)\cdot \sin^\frac{4}{2}(x)=\sin^\frac{5}{2}(x)\]

Answer 11

4/2=2

Answer 12

I understand that, can you just factor out the 2nd part? It becomes sqrt(sinx)*cosx (1-?...)

Answer 13

so that still means you don't know that:
\[\sin^\frac{1}{2}(x)\cdot \sin^\frac{4}{2}(x)=\sin^\frac{5}{2}(x)\]
?

Answer 14

\[\sin^\frac{1}{2}(x)\cos(x)-\sin^{\frac{5}{2}}\cos(x)\]
\[\sin^\frac{1}{2}(x)\cos(x)(1- ? )\]
If you understand what I'm saying about 1/2+4/2=5/2 then you should be able to fill in that blank

Answer 15

So it's \[(\sqrt{sinx} \times cosx) \times (1-\sin \frac{ 4 }{ 2 }) = \cos ^{3} \times \sqrt{sinx} ?\]

Answer 16

well you are missing the x's in some parts and that 4/2 is an exponent

Answer 17

and what is 1-sin^2(x)?

Answer 18

cos^2 (x) is I remeber

Answer 19

yep and cos(x)*cos^2(x) is the desired cos^3(x) that is on the other side of the equation

Answer 20

so it's solved? :O

Answer 21

no we proved both sides were the same
we didn't solve

Answer 22

well, that's what I meant, finding x values are a different story. (pi/4?) Thanks, I thought it would involving stating cosine in terms of sine, which would add more sqaure roots and throw me off further.