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# Is the trigonometry we are doing correct?

I am considering $\cos\theta=\sec\theta$then $\cos\theta=\frac{1}{\cos\theta}$ $=>\cos^2\theta=1$ $=>\cos\theta=1$ $=>\theta=0^\circ$here $\sin\theta=\sin 0^\circ=0.$ the next way again consider $\cos\theta=\sec\theta$ $=>\frac{adj}{hyp}=\frac{hyp}{adj}$ $=>\frac{adj^2}{hyp^2}=1$ $=>\cos^2\theta=1$ $=>\cos\theta=-1$ nothing too interesting here

Note by Sudoku Subbu
1 year, 9 months ago

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Note: $$\cos ^2 \theta = 1 \not \Rightarrow \cos \theta = 1$$. Do you see why? Staff · 1 year, 9 months ago

why staff iam sending the sqare to that side so $\cos\theta=1$ hmmm.. ttooo simple concept · 1 year, 9 months ago

If $$x^2 = 1$$, is $$x = 1$$ the only solution? Staff · 1 year, 9 months ago

ya ya i understand your concept if $x^2=1$ then x may be -1 or +1 thanks calvin i'll change · 1 year, 9 months ago

Right, that is a very common mistake made when trying to solve such equations.

The best approach would be to say $$0 = x^2 - 1 = (x-1)(x+1)$$ and hence $$x = 1, -1$$. Staff · 1 year, 9 months ago

thanks for your help visit my other notes and reply if there is any mistake · 1 year, 9 months ago

We can't say "concept" when we talk about "Fundamental theorem of algebra". · 1 year, 8 months ago

Remember the Pythagorean Identity: $\sin^2{\theta}+\cos^2{\theta}=1$ If $$\cos^2{\theta}=1$$, this implies that: $\sin^2{\theta}=0 \\ \sin{\theta}=0$ ¿For what angles is this true? For any angle of the form $\theta=n\pi$ As can be easily verified seeing any graph of the sine function. · 1 year, 9 months ago

Another thing... you don't just consider the first quadrant angle.. Consider also its coterminal angles.. · 1 year, 9 months ago