Trig identity (1 Viewer)

[philly2000]

Prove

Thanks!

 

[azureus88]

[maths]\sin^25a-\sin^23a[/maths]

[maths]=(\sin5a-\sin3a)(\sin5a+\sin2a)[/maths]

Using sums to products (which you can derive):

[maths]=(2\cos4a\sin a)(2\sin4a\cos a)[/maths]

[maths]=(2\sin4a\cos4a)(2\sin a\cos a)[/maths]

[maths]=\sin8a\sin2a[/maths]

 

[philly2000]

Is there another way to do it? I havent learnt the sums to products method.

 

[lolokay]

sums to products just means you break sin5a and sin3a into sin(4a + a) and sin(4a - a)

you could also use sin8Asin2A = sin(5a + 3a)sin(5a - 3a)

 

[azureus88]

i dont know of another method at the momment, but i guess you can go:

[maths]\sin8a\sin2a[/maths]

[maths]=(2\sin4a\cos4a)(2\sin a\cos a)[/maths]

[maths]=(2\cos4a\sin a)(2\sin4a\cos a)[/maths]

[maths]=(\sin(4a+a)-\sin(4a-a))(\sin(4a+a)+\sin(4a-a))[/maths]

[maths]=\sin^25a-\sin^23a[/maths]

pretty much same as sums/products

 

[Drongoski]

The way azureus88 did is the neatest. The sums to product is not hard (I feel it an oversight being left out of syllabus) and is derived from the sin(a + b) etc identities. Doing without is possible and is very similar to the one posted earlier in Mathematics ?

 

[philly2000]

thank you guys! i get it now

 

[studentcheese]

hehe I remember this question. Everyone in the class thought that Difference of two squares would solve it.

 

[youngminii]

I uhh, thought the sums to products method wasn't in the Syllabus anymore?