I think you can also prove it by: 1. Taking the square root of cos^4x-sin^4x 2. Using the Reduction formula for both Cos^2x and Sin^2x 3. Multiplying by 2 to eliminate the fraction 4. It should simplify to 2cos(2x) 5. Divide by 2 again should leave you with Cos(2x) on the left side.
hey i have a question how did u get 1? cuz its cos^2x(cos^2x) but how did u get it to equal 1??? i understand hiw u got (cos^2x-sin^2x) in the bracket but i dont understand how u get the 1
If you are "solving" for something, it must be in the form of an equation and you then solve for one of the variables. Are you asking if it can be written in a simplified form?
@@MichelvanBiezenactually teacher I don't want to “solve it “ I just wanted to write it as a product, such as you did in this video th-cam.com/video/G2w6B8bEd30/w-d-xo.html But when I tried to write it as a product I couldn't, because there is the (3) that is multiplicated with cos(2x)
@@MichelvanBiezen so teacher does it mean that there's is no way to write it using identities?? And is there any alternative way to make it as a product? ( this question is a homework from my teacher, and she said that there is a way to make it, but not a direct one so we need to think about it )
![PreCalculus - Trigonometry: Trig Identities (42 of 57) Prove [sinx+sin(5x)]/[cosx+cos(5x)]=tan3x](http://i.ytimg.com/vi/0yi6RnMZcM8/mqdefault.jpg)
![PreCalculus - Trigonometry: Trig Identities (42 of 57) Prove [sinx+sin(5x)]/[cosx+cos(5x)]=tan3x](/img/tr.png)
![ว้าว!ลุงบีมแบกทองมารับขวัญหลานกันยกบ้าน [cc] แดนแพทตี้ SS2 | EP.49 |](http://i.ytimg.com/vi/Osihq1OAs-4/mqdefault.jpg)
I think you can also prove it by:
1. Taking the square root of cos^4x-sin^4x
2. Using the Reduction formula for both Cos^2x and Sin^2x
3. Multiplying by 2 to eliminate the fraction
4. It should simplify to 2cos(2x)
5. Divide by 2 again should leave you with Cos(2x) on the left side.
+Dan Man Yes, there are often multiple methods to solve these types of problems. Thanks for sharing.
Great Videos! By the way, how would you write 3/4*sin(x)+1/4*sin(3x) as a product
Excellent!
Thank you.
That was perfect thank you
Thanks
Welcome
Dear Sir,
Any websites that allow us to discuss other questions?
Ngoi Kuan Binaeun
What type of questions?
thank you so much!!!!
All things so great._.!
0:32 "it's the difference of squares that can be factored" could you please show and explain that step?
Tony
The general form of factoring the difference of squares is:
a^2 - b^2 = (a - b) * (a+b)
+tony nkwaya yeah I'm sitting here like wait what lol
+tony nkwaya oops nevermind. I figured it out lol all logic and good ole common sense around these trig identities.
hey i have a question how did u get 1? cuz its cos^2x(cos^2x) but how did u get it to equal 1???
i understand hiw u got (cos^2x-sin^2x) in the bracket but i dont understand how u get the 1
+lebreezy james
cos^2x + sin^2x = 1
Ahh I remember that now thanks a lot
Thank u teacher
But what about
cos(4x) - 3cos(2x)
I couldn't solve it
If you are "solving" for something, it must be in the form of an equation and you then solve for one of the variables. Are you asking if it can be written in a simplified form?
@@MichelvanBiezenactually teacher I don't want to “solve it “ I just wanted to write it as a product, such as you did in this video th-cam.com/video/G2w6B8bEd30/w-d-xo.html
But when I tried to write it as a product I couldn't, because there is the (3) that is multiplicated with cos(2x)
So my question is how could I write it as a product
( and thank you for replying)
Because the "3" in front of the cos(2x) you will not be able to find any identities to converet it to.
@@MichelvanBiezen so teacher does it mean that there's is no way to write it using identities??
And is there any alternative way to make it as a product?
( this question is a homework from my teacher, and she said that there is a way to make it, but not a direct one so we need to think about it )
هذا السؤال جابته النا الست😢
We didn't quite get your comment.
Prove this pls
cos^2(4x) - 4cos^2(2x)sin^2(2x)
Shouldn't there be an equal sign somewhere?