Given cos (x-pai / 6) + SiNx = 4 times root 3 / 5, then sin (x + 7 PAI)=

Given cos (x-pai / 6) + SiNx = 4 times root 3 / 5, then sin (x + 7 PAI)=

cos(x-π/6)+sinx=4√3/5
cosxcosπ/6+sinxsinπ/6+sinx=4√3/5
(√3/2)cosx+(3/2)sinx=4√3/5
(1/2)cosx+(√3/2)sinx=4/5
sin(π/6+x)=4/5
sin(7π/6+x)=-4/5

1 + cosx + SiNx is not equal to o, sin (x + Pai / 4) is not equal to - radical 2 / 2, why,

sinx+cosx≠-1
√2/2sinx+√2/2cosx≠-√2/2
sinxcosπ/4+cosxsinπ/4≠-√2/2
sin(x+π/4)≠-√2/2

Simplify √ 1 + CoS / 1-cosx + √ 1-cosx / 1 + cosx X in the second quadrant. Is the final answer 2 / SiNx or - 2 / SiNx

2/SINX
The sine of the second quadrant is positive

1 if cosx = - 4 / 5 and X is the third quadrant angle, then sin (x + π / 4) = 2 SiNx = 2 / 3, find cos (π - 2x) = tomorrow Which 2 in the middle is the second question

1. X is the third quadrant angle, SiNx = - root sign (1-cosx * cosx) = 3 / 5sin (x + π / 4) = sinxcos (π / 4) + cosxsin (π / 4) = [root sign (2) / 2] [- 3 / 5-4 / 5] = - 7 root sign 2 / 102, cos (π - 2x) = - cos (π - 2x) = 2sinx * sinx-1 = 2 * (- 3 / 5) * (- 3 / 5) - 1 = - 7 / 25

If sin (x ^ 3) + cos (x ^ 3) = 1, find SiNx + cosx Urgent, the third power of degrees, Want process······

X = π (2n + 1 / 2) or 2n π
sinx+cosx=1
+Let's use the drawing method. Take x ^ 3 as a whole, and make an image of y = SiNx + cosx = root twice sin (x + π / 4) on (- ∞, + ∞), and an image of y = 1. Look at the abscissa corresponding to the intersection, find the corresponding x, and then find SiNx + cosx

Given SiNx + cosx = 1 / 5, X belongs to (0, π), find the value of COS ^ 3 x-sin ^ 3 X

SiNx = acosx = Ba + B = 1 / 5 (a + b) ^ 2 = a ^ 2 + B ^ 2 + 2Ab = 1 + 2Ab = 1 / 25ab = - 12 / 25 (B-A) ^ 2 = a ^ 2 + B ^ 2-2ab = 1-2 * (- 12 / 25) = 49 / 25b-a = - 7 / 5 or 7 / 5B ^ 3-A ^ 3 = (B-A) * (b ^ 2 + AB) = (B-A) * (1 + AB) = 7 / 5 * (1-12 / 25) = 91 / 125 or - 91 / 125