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

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TopNewestNote: \( \cos ^2 \theta = 1 \not \Rightarrow \cos \theta = 1 \). Do you see why? – Calvin Lin Staff · 1 year, 9 months ago

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– Sudoku Subbu · 1 year, 9 months ago

why staff iam sending the sqare to that side so \[\cos\theta=1\] hmmm.. ttooo simple conceptLog in to reply

– Calvin Lin Staff · 1 year, 9 months ago

If \( x^2 = 1 \), is \( x = 1 \) the only solution?Log in to reply

– Sudoku Subbu · 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 changeLog in to reply

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

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– Sudoku Subbu · 1 year, 9 months ago

thanks for your help visit my other notes and reply if there is any mistakeLog in to reply

– Andi Popescu · 1 year, 8 months ago

We can't say "concept" when we talk about "Fundamental theorem of algebra".Log in to reply

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. – Ricardo Adán Cortés Martín · 1 year, 9 months ago

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

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boss what r u trying to tel?? – Saran .P.S · 1 year ago

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