
How do you verify cos^4x - sin^4x = 1 - 2sin^2x? - Socratic
Mar 19, 2018 · To prove that #cos^4x-sin^4x=1-2sin^2x#, we'll need the Pythagorean identity and a variation on the Pythagorean identity:
How do you simplify sin4theta to trigonometric functions of a unit ...
Feb 13, 2016 · How do you use a double-angle identity to find the exact value of sin 120°? How do you use double angle identities to solve equations? How do you find all solutions for #sin 2x …
How do you find the exact values #sin (pi/4)# using the ... - Socratic
Aug 4, 2015 · How do you find the exact values #sin (pi/4)# using the special triangles?
Sin^4x -cos^4x= cos3x Could you solve this? - Socratic
Apr 5, 2018 · #sin ^4 x - cos ^4 x = cos 3x # #(sin ^2 x + cos ^2 x)(sin ^2 x - cos ^2 x ) = cos 3x # #-cos 2x = cos 3x # #cos (180^circ - 2x) = cos 3x # I've been on Socratic for a couple of …
How do I simplify sin^4x-2sin^2x+1? - Socratic
Mar 12, 2018 · It is a perfect square: sin^4(x)-2sin^2(x)+1= (sin^2(x)-1)^2. What are the units used for the ideal gas law
How do you express sin^4theta-cos^3theta in terms of non
Feb 7, 2016 · If we use sin^4x as an example: If you are versed in complex numbers, remember that: cos(nx)+isin(nx)=e^(i nx) From this it can be shown: sin(nx) = (e^(i nx)-e^(-i nx))/(2i) and …
How do you graph # y=sin(4/3x)#? - Socratic
Jun 17, 2018 · When graphing trigonometric functions, it's important to identify the amplitude and period from the equation provided. For example, the sine graph: y=asin(bx) has amplitude a …
How do you use csctheta=4 to find costheta? | Socratic
Jan 15, 2017 · cos theta=0.968245836 cosec theta=4 the opposite of cosec theta= sin theta sin theta= 0.25 theta =14°.47751219 = 14°28'39" theta lies in the first quadrant where sin and …
What is the exact vaule for #sin((3pi)/4)# and #sin((9pi)/4)
Apr 21, 2015 · sin(3/4 pi) =(sqrt(2))/2 and sin(9/4 pi) =(sqrt(2))/2 Solution: You solve such problems by using the reduction formulas for trigonometric functions.
Sin^6(π/8)+sin^6(3π/8)+sin^6(5π/8)+sin^6(7π/8)=5/4? - Socratic
Jun 16, 2018 · #LHS=Sin^6(π/8)+sin^6(3π/8)+sin^6(5π/8)+sin^6(7π/8)# #=Sin^6(π/8)+sin^6((3π)/8)+sin^6(pi-(3π)/8)+sin^6(pi-π/8)#