Help trig
I don’t get how they did part a and b…
For part a how do they do it I’m so lost 😭
it’s q 7 btw 
For part a how do they do it I’m so lost 😭
(edited 11 months ago)
Use the identity sin^2x + cos^2x = 1
Then rearrange to get cos^2x = 1 - sin^2x and sub this in for cos^2x
Factorise out the minus to get 2-
Then factorise inside brackets to get (sin x- 1)^2
Let me know if this makes sense!!
Then rearrange to get cos^2x = 1 - sin^2x and sub this in for cos^2x
Factorise out the minus to get 2-
Then factorise inside brackets to get (sin x- 1)^2
Let me know if this makes sense!!
I'm not 100% sure, but I think the question just puts in a +2 by taking 2 away from the 1 in the quadratic.
- So you have the initial cos(x)^2 + 2 sin(x)
- This becomes 1 - sin(x)^2 + 2 sin(x) after applying the trig identity
- Then you put in the 2 at the front by taking away 2 from the 1 you have from above which gives you 2 + (-1) - sin(x)^2 + 2 sin(x)
- If you keep the 2 outside and add in a (-), you end up with 2 - (1 + sin(x)^2 - 2 sin(x)) as taking a (-) out swaps the sign for everything in the bracket
- You have a normal quadratic in the bracket which looks like b^2 - 2b + 1 which factorises into (b-1)^2 where I used b as a substitute for sin(x)
- You put this now factorised quadratic back into the equation to give yourself 2 - (sin(x) - 1)^2 like shown in the question
That's the method I would use. Again, I'm not sure if it's right though since the 2 at the front out of nowhere is kind of throwing me, and I'm guessing this is how it'd be done.
- So you have the initial cos(x)^2 + 2 sin(x)
- This becomes 1 - sin(x)^2 + 2 sin(x) after applying the trig identity
- Then you put in the 2 at the front by taking away 2 from the 1 you have from above which gives you 2 + (-1) - sin(x)^2 + 2 sin(x)
- If you keep the 2 outside and add in a (-), you end up with 2 - (1 + sin(x)^2 - 2 sin(x)) as taking a (-) out swaps the sign for everything in the bracket
- You have a normal quadratic in the bracket which looks like b^2 - 2b + 1 which factorises into (b-1)^2 where I used b as a substitute for sin(x)
- You put this now factorised quadratic back into the equation to give yourself 2 - (sin(x) - 1)^2 like shown in the question
That's the method I would use. Again, I'm not sure if it's right though since the 2 at the front out of nowhere is kind of throwing me, and I'm guessing this is how it'd be done.
Original post by lkrnix
Use the identity sin^2x + cos^2x = 1
Then rearrange to get cos^2x = 1 - sin^2x and sub this in for cos^2x
Factorise out the minus to get 2-
Then factorise inside brackets to get (sin x- 1)^2
Let me know if this makes sense!!
Then rearrange to get cos^2x = 1 - sin^2x and sub this in for cos^2x
Factorise out the minus to get 2-
Then factorise inside brackets to get (sin x- 1)^2
Let me know if this makes sense!!
It’s makes no sense sorry I’m so dumb 😭
move for my finals in 5 days and I’m screwed I can’t remember anything and I’m having mental breakdowns all the time 😭
Original post by Alevelhelp.1
It’s makes no sense sorry I’m so dumb 😭
move for my finals in 5 days and I’m screwed I can’t remember anything and I’m having mental breakdowns all the time 😭
move for my finals in 5 days and I’m screwed I can’t remember anything and I’m having mental breakdowns all the time 😭
As an alternative to the above, you can start with the final form of the expression they require so just say:
Consider 2 - (sinx - 1)^2. This = 2 - (sin^2x - 2sinx + 1) = 2 - sin^2x + 2sinx - 1. You now have a term in six that you want to keep, two constants you want to combine and a -sin^2x. Now you can use your standard trig identity and show that this final expression is identical to the original form of f(x) given.
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