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Solve it plz......

Cos x + root 3 sin x =root 2

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Cos x + root 3 sin x =root 2
cos x + ö3 * Sin x = ö2
squaring both the side
(cos x + ö3 * Sin x)2 = (ö2)2
Cos2 x + 3 * Sin2 x = 2
Cos2 x + Sin2 x + 2 * Sin2 x = 2
1 + 2 * Sin2 x= 2
2 * Sin2 x = 2-1
2 * Sin2 x = 1
Sin2 x = ½
Sin x = ö½
Sin x = 1/V2= Sin 45
X = 450

Posted on Jan 10, 2009

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How can I solve Sinx+cosx=7/5?


Square each side

Sin ^2 (x) + cos ^2 (x) +2 sin (x)cos (x) = 49/25

1 + 2sin (x)cos (x) = 1.960

sin (2x) = 0.960

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Press the relevant function key [SIN],[COS],[TAN] followed by the angle value.
For inverse trigonometric functions [SIN^-1], COS^-1], or [TAN^-1], Press [SHIFT][SIN], [SHIFT][COS], or [SHIFT][TAN].
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1 Answer

Differentiate each of the following w.r.t.x; 29.sin2xsinx


Use the rule for differentiating products of functions: ()' signifies derivative
(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'
But
  1. (29)'=0 derivative of a constant is zero
  2. (sin(2X))'=cos(2X)*(2X)'=2*cos(2X)
  3. (sin(X))'=cos(X)
Result is
(29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)

You could also have cast your formula in the form
sin(2X)*sin(X)= 1/2[ cos(2X-X)-cos(2X+X)]=1/2[cos(X)-cos(3X)]
then calculated the derivative of
29/2*[cos(X)-cos(3X)]
which is
29/2*[-si(X) +3*sin(3X)]

The challenge for you is to prove that the two forms are equivalent
29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)=29/2*[-si(X) +3*sin(3X)]

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1 Answer

Sin x cos x = -1/2 solve for x


sin x cos x = -1/2
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If f(x) = sin x, what is the 99th derivative of f(x)?


Hello,
Frankly my dear, who wants to compute the 99th derivative of sin(x)?

d/dx(sin(x))= cos(x)
d^2/dx^2 (sin(x))= d/dx( cos(x))=-sin(x)
d^3/dx^3 (sin(x))= -cos(x)
d^4/dx^4 (sin(x))=d/dx(-cos(x))=sinx.
In 100 successive differentiations you perform the foregoing sequence 100/4=25 times, and you end up with sin(x). Hence in the 99th step the derivative was -cos(x).

Hope it helps.

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1 Answer

(1+cotx-cosecx)(1+tanx+secx)=2


I shall attempt :D
1) cosec A + cot A = 3
we know that (cot A)^2 + 1 = (cosec A)^2
Hence, (cosec A)^2 - (cot A)^2 = 1
thus, (cosec A + cot A) (cosec A - cot A) = 1
3 (cosec A - cot A) = 1
(cosec A - cot A) = 1/3

(cosec A - cot A) = 1/3
(cosec A + cot A) = 3
Summing them, 2 cosec A = 3 1/3
cosec A = 6 2/3 = 5/3
sin A = 0.15
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2) Prove that (1+tan x - sec x)(1 + cot x + cosec x) =2
expand
LHS= 1 + cot x + cosec x + tan x + 1 + tan x cosec x - sec x - sec x cot x - sec x cosec x
We can calculate that
tan x cosec x = sec x (since tan x = sin x / cos x)
sec x cot x = cosec x
so the above is
LHS = 1 + cot x + cosec x + tan x + 1 + sec x - sec x - cosec x - sec x cosec x
LHS = 2 + cot x + tan x - sec x cosec x
LHS = 2 + cos x / sin x + sin x / cos x - 1 / (sin x cos x)
LHS = 2 + [{cos x}^2 + {sin x}^2 - 1] / (sin x cos x)
LHS = 2 (proved)

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