Trigonometry Problem: Finding Length of Vector a = (-2sint, 0, -2cost)

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Homework Statement


a = 1/sqrt(29) * (-2*sint, 0, -2*cost)
length of a is?

Homework Equations





The Attempt at a Solution


ignoring 0, we have sqrt{[(-2*sint)/sqrt(29)]^2 + [(-2*cost)/sqrt(29)]^2}. I get 2/sqrt(29) but answer is [2*sqrt(2)]/sqrt(29). Can anyone do this step by step and show me please..
 
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hanelliot said:

Homework Statement


a = 1/sqrt(29) * (-2*sint, 0, -2*cost)
length of a is?

Homework Equations





The Attempt at a Solution


ignoring 0, we have sqrt{[(-2*sint)/sqrt(29)]^2 + [(-2*cost)/sqrt(29)]^2}. I get 2/sqrt(29) but answer is [2*sqrt(2)]/sqrt(29). Can anyone do this step by step and show me please..
No need to. You answer is correct. 2\sqrt{2}/\sqrt{29} would be \sqrt{4+ 4}/\sqrt{29} but 4sin^2(t)+ 4cos^2(t)= 4 not 4+ 4.
 
hmm so solution must be wrong.. I wonder why no one pointed it out tho
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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