3x + 4y = 7 5x + 10y = 15 If you were going to solve th … e system above, what would be the best first step according to the strategy you just learned?cos4(x)-sin4(x). Simplify the expression. Tap for more steps... cos2(x)-sin2(x). Apply the cosine double-angle identity.How do you use the half angle identity to find cos 105? How do you find the exact value for #sin105# using the half‐angle identity?we know from identity that : 2sin(x).cos(x) = sin(2x). for max value of y second term must be min. which is when sin(2x) is 0, so max value is 1.`int sin^4 x*cos^4 x dx= int (1 - 2cos^2 2x + cos^4 2x)^2/16 dx`. Using the linearity property of integrals yields What is the integral `int cos^5 x sin^4 x dx`. 2 Educator answers.
Simplify cos(x)^4-sin(x)^4 | Mathway
How do you verify cos4x−sin4x(2cos2x−1)2 ? cos2x Explanation: Knowing that: cos2x=cos2x−sin2x=2cos2x−1 and sin2x+cos2x=1Решите уравнение: sin^4x+cos^4x=sinxcosx....cos x +i \sin x)^4$. Then I use the binomial theorem to expand this fourth power, and comparing real and imaginary parts, I conclude that $\begingroup$ See also: How to simplify $\sin^4 x+\cos^4 x$ using trigonometrical identities? $\endgroup...integral of sin^4 (x) * cos^4 (x) I tried a bunch of u substitutions but nothing so far has worked.
How do you simplify cos^4x-sin^4x? | Socratic
Click hereto get an answer to your question(cos^4x - sin^4x) is equal to. (cos4x−sin4x) is equal to. Answer. As we know that.Now, there are several ways of simplifying this trigonometric expression depending on whether you want it to be in terms of sin x, cos x, tan x or multiple angles. 1st option[math]\sin{x}=\pm\frac{\sqrt{2}}{2}[/math]. 22x\hfill\\\cos^22x=0\hfill\\\end{gathered}[/math].Задача 14267 cos4x-sin4x=1... Условие. cos4x-sin4x=1. математика 10-11 класс 2599.1 - 2cos^2(x) + cos^4(x) factor and use the fundamental identities to simplify. How to differentiate y=sin(4x) using the Chain Rule.
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$\begingroup$Derive the id $\cos^4 x + \sin^4 x=\frac14 \cos (4x) +\frac34$
I do know $e^i4x=\cos (4x) + i \sin (4x)=(\cos x +i \sin x)^4$. Then I exploit the binomial theorem to increase this fourth energy, and comparing actual and imaginary parts, I conclude that $\cos^4 x + \sin^4 x = \cos (4x) + 6 \cos^2 (x) \sin^2 (x)$.
So now I wish to display that $\cos (4x) + 6 \cos^2 (x) \sin^2 (x)=\frac14 \cos (4x) +\frac34$, which has stumped me.
asked Feb 17 '15 at 15:40
DuckyDucky1,8541616 silver badges2929 bronze badges
$\endgroup$ 1 $\begingroup$$\cos^4x+\sin^4 x\=(\cos^2x+\sin^2x)^2-2(\sin x\cos x)^2\=1-2(\frac12\sin(2x))^2\=1-\frac12\sin^2(2x)\=1-\frac12\frac1-\cos(4x)2\=\frac14\cos(4x)+\frac34$
spoke back Feb 17 '15 at 15:45
UncountableUncountable3,33277 silver badges1717 bronze badges
$\endgroup$ 1 $\begingroup$$$\cos^4x+\sin^4x\=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x\=1-\frac12\sin^22x=1-\frac12\left(\frac12(1-\cos4x)\proper)=\frac34+\frac14\cos4x$$ As: $$\sin^2x+\cos^2x=1;\(a+b)^2=a^2+b^2+2ab;\\sin2x=2\sin x\cos x;\\cos2x=\cos^2x-\sin^2x=1-2\sin^2x$$
Also: $$\cos^4x+\sin^4x=\left(\frace^ix+e^-ix2\proper)^4+\left(\frace^ix-e^-ix2\right)^4\=\frac116(\small e^-4 ix+4 e^-2 ix+4 e^2 ix+e^4 ix+6+e^-4 ix-4 e^-2 ix-4 e^2 ix+e^4i x+6)\=\frac34+\frac14\left(\frace^4ix+e^-4ix2\right)=\frac34+\frac14\cos4x$$
responded Feb 17 '15 at 15:44
RE60KRE60K17k22 gold badges2727 silver badges6969 bronze badges
$\endgroup$ 3 $\begingroup$$$\beginalign \cos^4x + \sin^4 x &= \left( \cos^2 x \right)^2 + \left( \sin^2 x \proper)^2 \[4pt] &= \left( \frac1 + \cos 2 x2\right)^2 + \left( \frac1-\cos 2x2 \right)^2 \[4pt] &= \frac12\left( 1 + \cos^2 2 x \proper) \[4pt] &= \frac12\left( 1 + \frac1+\cos 4 x2 \right) \[4pt] &= \frac14\left(\; 3 + \cos 4 x \;\proper) \endalign$$
From your particular stopping place, you'll want to proceed thusly: $$\startalign \cos 4 x + 6 \sin^2 x \cos^2 x &= \cos 4 x + \frac32\cdot(2\sin x \cos x)^2 \[4pt] &= \cos 4 x + \frac32\cdot \sin^2 2 x \[4pt] &= \cos 4 x + \frac32\cdot \frac1 - \cos 4 x2 \[4pt] &= \frac14\left(\;3 + \cos 4 x\;\right) \finishalign$$
answered Feb 17 '15 at 15:51
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$\endgroup$ $\begingroup$$\cos^2 x \sin^2 x = (1 - \sin^2 x) \sin^ 2 x = \sin^2 x - \sin^ 4 x$ and by means of the same argument $\cos^2 x \sin^2 x = \cos^2 x - \cos^ 4 x$. Take the average of those two identities to acquire $$ \cos^2 x \sin^2 x = \frac12 - \frac\cos^4x + \sin^4 x2 \, . $$ Now exchange this and simplify.
answered Feb 17 '15 at 15:45
Hans EnglerHans Engler12.8k22 gold badges2323 silver badges3838 bronze badges
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