...text book gives a different answer

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lihinii

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Solve cos(inverse)x - sin(inverse)x = pi/6.

Let cos(in)x = A cos A =x , sin A = sq.rt.(1- x sq)
sin(in)x = B sin B =x , cos B = sq.rt.(1-x sq)

A -B = pi/6
cos A cos B + sin A sin B = cos pi/6 = (sq.rt.3)/2
x sq.rt.( 1-x sq) + sq.rt.(1-x sq) x = (sq.rt.3)/2
2x sq rt.( 1-x sq) = (sq.rt. 3)/2
4xsq(1-x sq) = 3/4
x = + or - 1/2 x + or - (sq rt 3)/2

When someone has time can please check this....coz a text book gives a differfnt answer.
Thanks so much
 
lihinii said:
Code: 
Solve  cos(inverse)x - sin(inverse)x = pi/6.

   Let cos(in)x = A       cos A =x ,  sin A = sq.rt.(1- x sq)
         sin(in)x = B        sin B =x ,  cos B = sq.rt.(1-x sq)
      
         A -B =  pi/6
         cos A cos B + sin A sin B = cos pi/6 = (sq.rt.3)/2
         x sq.rt.( 1-x sq)  + sq.rt.(1-x sq) x = (sq.rt.3)/2
         2x sq rt.( 1-x sq) = (sq.rt. 3)/2
         4xsq(1-x sq) = 3/4
           x = + or - 1/2            x + or - (sq rt 3)/2

You need to use the Code tags, if you want to space stuff apart. As you can see above, even with Code tags, the columns do not line up. That's because you need to use a fixed-width font when viewing your "spacing". To fix this, use the [Preview] button to see exactly where spacing adjustments need to be made.

:idea: OR, don't try to design columns. Just type your work vertically, one column after another. That's easiest.

You need to check your own answers. I mean, one of the four candidates that you got is correct, and the other three are not.

Evaluate the given equation for these four values of x. Find the one that works.

EG: x = sqrt(3)/2

arccos[sqrt(3)/2] - arcsin[sqrt(3)/2]

Pi/6 - Pi/3 = -Pi/6

No good.
Click to expand...
 
One way;\displaystyle One \ way;

arccos(x)−arcsin(x) = π6, let θ = arccos(x)   ⟹   cos(θ) = x.\displaystyle arccos(x)-arcsin(x) \ = \ \frac{\pi}{6}, \ let \ \theta \ = \ arccos(x) \ \implies \ cos(\theta) \ = \ x.

Ergo, θ−arcsin(x) = π6   ⟹   arcsin(x) = (θ−π6),   ⟹   sin(θ−π6) = x\displaystyle Ergo, \ \theta-arcsin(x) \ = \ \frac{\pi}{6} \ \implies \ arcsin(x) \ = \ \bigg(\theta-\frac{\pi}{6}\bigg), \ \implies \ sin\bigg(\theta-\frac{\pi}{6}\bigg) \ = \ x

Hence, sin(θ−π6) = cos(θ)\displaystyle Hence, \ sin\bigg(\theta-\frac{\pi}{6}\bigg) \ = \ cos(\theta)

Therefore, sin(θ)cos(π/6)−cos(θ)sin(π/6) = cos(θ)\displaystyle Therefore, \ sin(\theta)cos(\pi/6)-cos(\theta)sin(\pi/6) \ = \ cos(\theta)

= 3sin(θ)2−cos(θ)2 = cos(θ)   ⟹   3sin(θ)2 = 3cos(θ)2\displaystyle = \ \frac{\sqrt3 sin(\theta)}{2}-\frac{cos(\theta)}{2} \ = \ cos(\theta) \ \implies \ \frac{\sqrt3 sin(\theta)}{2} \ = \ \frac{3cos(\theta)}{2}

= 3sin(θ) = 3cos(θ)   ⟹   sin(θ) = 3cos(θ)3   ⟹   tan(θ) = 33, θ = π3\displaystyle = \ \sqrt3sin(\theta) \ = \ 3cos(\theta) \ \implies \ sin(\theta) \ = \ \frac{3cos(\theta)}{\sqrt3} \ \implies \ tan(\theta) \ = \ \frac{3}{\sqrt3}, \ \theta \ = \ \frac{\pi}{3}

Ergo, cos(π3) = x, x = 12, QED.\displaystyle Ergo, \ cos\bigg(\frac{\pi}{3}\bigg) \ = \ x, \ x \ = \ \frac{1}{2}, \ QED.
 
Hello, lihinii!

Yet another approach . . .

cos⁡-1(x)  and  sin⁡-1(x)  are complementary angles!\displaystyle \cos^{\text{-}1}(x)\,\text{ and }\,\sin^{\text{-}1}(x)\,\text{ are }complementary\text{ angles!}

Solve:   cos⁡-1(x)−sin⁡-1(x) = π6\displaystyle \text{Solve: }\;\cos^{\text{-}1}(x) - \sin^{\text{-}1}(x) \:=\: \frac{\pi}{6} .[1]
Click to expand...

Let  A = cos⁡−1(x)\displaystyle \text{Let }\,A \,=\, \cos^{-1}(x) .[2]

. . Then:  cos⁡A = x\displaystyle \text{Then: }\,\cos A \:=\:x

We have:  cos⁡A = x1 = adjhyp\displaystyle \text{We have: }\:\cos A \:=\:\frac{x}{1} \:=\:\frac{adj}{hyp}


A is in a right triangle with:  adj=x,  hyp=1.\displaystyle A\text{ is in a right triangle with: }\,adj = x,\;hyp = 1.

Code: 
                        * B
                     *  *
              1   *     *
               *        *
            *           *
         *              *
    A *  *  *  *  *  *  * C
               x

We can see that:  sin⁡B = x    where B = π2−A\displaystyle \text{We can see that: }\:\sin B \:=\:x\,\;\text{ where }B \:=\:\frac{\pi}{2} - A

So we have:  sin⁡(π2−A) = x⇒π2−A = sin⁡-1(x)\displaystyle \text{So we have: }\:\sin\left(\frac{\pi}{2}-A\right) \:=\:x \quad\Rightarrow\quad \frac{\pi}{2} - A \:=\:\sin^{\text{-}1}(x) .[3]


Substitute [2] and [3] into [1]:

. . A−(π2−A) = π6⇒A−π2+A = π6\displaystyle A - \left(\frac{\pi}{2} - A\right) \:=\:\frac{\pi}{6} \quad\Rightarrow\quad A - \frac{\pi}{2} + A \:=\:\frac{\pi}{6}

. . 2A = 2π3⇒A = π3\displaystyle 2A \:=\:\frac{2\pi}{3} \quad\Rightarrow\quad A \:=\:\frac{\pi}{3}


Substitute into [2]:

. . π3 = cos⁡-1(x)⇒cos⁡(π3) = x⇒x = 12\displaystyle \frac{\pi}{3} \:=\:\cos^{\text{-}1}(x) \quad\Rightarrow\quad \cos\left(\frac{\pi}{3}\right) \:=\:x \quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{2}}
 
Thanks so much for that. But can you point out the error in my calculation.

Thanks
 
lihinii said:
Solve cos(inverse)x - sin(inverse)x = pi/6.

Let cos(in)x = A cos A =x , sin A = sq.rt.(1- x sq)
sin(in)x = B sin B =x , cos B = sq.rt.(1-x sq)

A -B = pi/6
cos A cos B + sin A sin B = cos pi/6 = (sq.rt.3)/2
x sq.rt.( 1-x sq) + sq.rt.(1-x sq) x = (sq.rt.3)/2
2x sq rt.( 1-x sq) = (sq.rt. 3)/2
4xsq(1-x sq) = 3/4
x = + or - 1/2 x + or - (sq rt 3)/2

When someone has time can please check this....coz a text book gives a differfnt answer.
Thanks so much
Click to expand...

You have four solutions - one of those satisfy your original equation and correct. Others are extraneous solutions - come up due to squaring terms.
 
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