How can you simplify [math]\cos^6x + \sin^6x [/math] to [math]1- 3\sin^2x \cos^2x ?[/math]
Well, its very easy if you know some basic formulas from algebra.[math]a^3 + b^3[/math][math]= (a+b)(a^2 + b^2 - ab)[/math][math]=(a+b)((a+b)^2 - 2ab - ab)[/math][math]=(a+b)^3 - 3ab(a+b)[/math]now substitute a and b as [math]a = \sin^2x, b = \cos^2x[/math]and use the identity [math]\sin^2x + cos^2x = 1[/math]You will get your answer which is[math]\sin^6x + \cos^6x = 1 - 3\sin^2x\cos^2x[/math]
How do you simplify cosx-cos^3x?
cos x - cos ^3 x cos x [ 1 - cos^2 x ] cos x [ sin^2 x] cos x sin^2 x
How do you simplify (cos^2x / sinx) + sinx?
.. cos^2(x) / sin(x) + sin(x) = [ 1 - sin^2(x) ] / sin(x) + sin(x) = 1 / sin(x) - sin(x) + sin(x) = 1 / sin(x)
How do I simplify x= sinx^-1 (-1/3) and x= sin^-1(2)?
Think what both statments mean, you wer solving for x but reality, you have 1/3=sin(x). the idea is you want a some kind of x where taking a sin of it will give you 1/3. So rembering defentions you remember that for some value x, sin(x)=oppsite/hypotesis. We can draw a picture to help think through the problem. So you want to find missing side first, using pythagarian theorm [math] a^2+b^2=c^2,[/math] substitute, [math]a^2+1=9 [/math], [math] a^2=8[/math], [math]a=2sqrt(2)[/math]. Now you definatly know the angle you are looking for is smallest, it is not is 30, 60,90 triangle(it doesn't always work out that simple) but we are looking it is opposite of the smallest side in this case use a calculator and use similar method as you would for 30 degreesex x=30, 210, n*360+30, n*360+210 (deegrees)for n=1, 2,3,4... the reason is we are looking for all the values that satisfy the original equation. The second problem is easier if you asking about 1/2=sin(x) is for 45, 45, 90 triangle and that is even simpler. so x=n*360+45, n*360+215 for n=0,1,2,3,4....
How to simplify sin2xsin3x?
cos(3x-2x) = cos3xcos2x + sin3xsin2x cos(3x+2x) = cos3xcos2x - sin3xsin2x cos(3x-2x) - cos(3x+2x) = 2sin3xsin2x sin2xsin3x = (1/2)[cos(3x-2x) - cos(3x+2x)] =(1/2)[cos(x) - cos(5x)] I am not sure of this is what you want but I hope it helps.
How can I simplify sinx-cosx+sin (x+pi/2)-cos (x+3pi/2)?
First we have to calculate value of sin(x+pi/2):-Formula :sin(a+b)=sin(a)cos(b)+cos(a)sin(b)=> sin(π/2+x)=sin(π/2)⋅cos(x)+cos(π/2)⋅sin(x)But sin(π/2)=1 and cos(π/2)=0So we have :=> sin(π2+x)=cos(x)Now we have to calculate cos(x+3pi/2):-Formula :cos(α+β)=cosαcosβ−sinαsinβ=> cos(x+3π/2)=cos(3π/2)cos(x)−sin(3π/2)sin(x)But Cos(3pi/2)= 0 and sin(3pi/2) = -1So we have=> Cos(x+3pi/2) = sin(x)Now come to the question= sin(x)-cos(x)+sin (x+pi/2)-cos (x+3pi/2)Put about calculated values= sin(x)-cos(x)+cos(x)-sin(x)= 0
How do I add and simplify (cosx/sinx)+(sinx/cosx)?
its easy. just make a common denominator of sinxcosx (cos^2(x)+sin^2(x))/sinxcosx the numerator is = to 1. (look at the trig identities. sin^2x+cos^2x=1. so you end up with (1)/(sinxcosx) which is just 1 over sine(x) times 1 over cos(x) you end up with Csc(x)Sec(x)
Simplify the expression cosx/tan^2x - cosx/sin^2x?
= cos x( 1/ tan ^2 x - 1/ sin ^2 x) = cos x( cot^2 x - cosec^2 x) = cos x( - 1) as cosec^2 x = 1+ cot^2 x = - cos x
How do I simplify the expression [math]\cos(\sin^{-1}(x))[/math] with a right triangle?
Thanks for the A2A!The inverse sine is the angle whose sine is equal to the parameter given into the inverse sine, so:Just pretend it’s a right triangle, okay? Sorry for my terrible handwriting.In this diagram, [math]\theta=\arcsin{x}[/math]. Using the Pythagorean Theorem, we can find the missing side:And so:[math]\cos{\arcsin{x}}=\cos{\theta}=\frac{\sqrt{1-x^2}}{1}=\sqrt{1-x^2}[/math]