Given sin (3 π - x) - cos (5 π + x) = 1 / 2 - (3 under the root sign) / 2, where x belongs to (0, π), find the values of SiN x, cos X and X

Given sin (3 π - x) - cos (5 π + x) = 1 / 2 - (3 under the root sign) / 2, where x belongs to (0, π), find the values of SiN x, cos X and X

Sin (3 π - x) - cos (5 π + x) = 1 / 2-sqr (3) / 2sinx + conx = 1 / 2-sqr (3) / 2 the square of both sides gets 1 + 2sinxconx = 1-sqr (3) / 2sin2x = - SQR (3) / 2 because x belongs to (0, π), 2x belongs to (0,2 π), so 2x = π / 3 + π or 2x = 2 π - π / 3x = 2 π / 3 x = 5 π / 6, so both SiNx conx can

Given SiNx + cosx = 2 / 3, find the value of sin ^ 4 + cos ^ 4

SiNx + cosx = 2 / 3. Sin is obtained when both sides are squared at the same time ² x+2sinxcosx+cos ² x=4/91+2sinxcosx=4/92sinxcosx=-5/9sinxcosx=-5/18(sinx)^4+(cosx)^4=[(sinx) ²+ (cosx) ²]²- 2(sinxcosx) ²= one ²- two × (-5/18) ²=...

Sin (sin (SiNx)) derivation

Let u = SiNx, u '= cosx
Let v = sin (SiNx) = sinu, V '= COSU * u' = cos (SiNx) * cosx
y=sin(sin(sinx))=sinv,y'=cosv*v'=cos(sin(sinx))*cos(sinx)*cosx

What is the derivative of sin ^ 3 (4x)?

Let u = 4x, then sin ^ 3 (4x) = sin ^ 3 (U); d[sin^3(4x)]/dx={d[sin^3(4x)]/du}{du/dx}={d[sin^3(u)]/du}{du/dx}=3sin^2(u)·cos(u)·{d(4x)/dx}=4·3sin^2(u)·cos(u)=12sin^2(4x)·cos4x)

What derivative = sin ^ 4x

According to the angle doubling formula cos2x = 1-2 (SiNx) ^ 2 = 2 (cosx) ^ 2-1 ((SiNx) ^ 2) ^ 2 = (1-cos2x) ^ 2 / 4 = 1 / 4 + [(cos2x) ^ 2] / 4-1 / 2cos2x = 1 / 4 + (1 + cos4x) / 8-1 / 2cos2x = 3 / 8 + 1 / 8cos4x-1 / 2cos2x, calculate the indefinite integral of the above formula to obtain 3x / 8 + 1 / 32sin4x-1 / 4sin2x + C, i.e. 3x / 8 + 1 / 32sin4

Y = sin ^ 4x / 4 + cos ^ 4x / 4 derivation

y=sin⁴x /4+cos⁴x /4
=(1/4)(sin⁴x+cos⁴x)
=(1/4)(sin⁴x+2sin ² xcos ² x+cos⁴x)-sin ² xcos ² x/2
=(1/4)(sin ² x+cos ² x) ²- (2sinxcosx) ²/ eight
=1/4-sin ² (2x)/8
=(-1/16)[2sin ² (2x)+1]+5/16
=(-1/16)cos(2x)+5/16
y'=(1/16)sin(2x)(2x)'=sin(2x)/8
Tip: this problem is simplified with trigonometric function knowledge, and then derivative, which is relatively simple