Is the trigonometry we are doing correct?

I am considering cos⁡θ=sec⁡θ\cos\theta=\sec\thetathen cos⁡θ=1cos⁡θ\cos\theta=\frac{1}{\cos\theta} =>cos⁡2θ=1=>\cos^2\theta=1 =>cos⁡θ=1=>\cos\theta=1 =>θ=0∘=>\theta=0^\circhere sin⁡θ=sin⁡0∘=0.\sin\theta=\sin 0^\circ=0. the next way again consider cos⁡θ=sec⁡θ\cos\theta=\sec\theta =>adjhyp=hypadj=>\frac{adj}{hyp}=\frac{hyp}{adj} =>adj2hyp2=1=>\frac{adj^2}{hyp^2}=1 =>cos⁡2θ=1=>\cos^2\theta=1 =>cos⁡θ=−1=>\cos\theta=-1 nothing too interesting here

#NumberTheory

Note by Sudoku Subbu
4 years, 6 months ago

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Note: cos⁡2θ=1⇏cos⁡θ=1 \cos ^2 \theta = 1 \not \Rightarrow \cos \theta = 1 . Do you see why?

Calvin Lin Staff - 4 years, 6 months ago

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why staff iam sending the sqare to that side so cos⁡θ=1\cos\theta=1 hmmm.. ttooo simple concept

sudoku subbu - 4 years, 6 months ago

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If x2=1 x^2 = 1 , is x=1 x = 1 the only solution?

Calvin Lin Staff - 4 years, 6 months ago

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@Calvin Lin – ya ya i understand your concept if x2=1x^2=1 then x may be -1 or +1 thanks calvin i'll change

sudoku subbu - 4 years, 6 months ago

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

The best approach would be to say 0=x2−1=(x−1)(x+1) 0 = x^2 - 1 = (x-1)(x+1) and hence x=1,−1 x = 1, -1 .

Calvin Lin Staff - 4 years, 6 months ago

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@Calvin Lin – thanks for your help visit my other notes and reply if there is any mistake

sudoku subbu - 4 years, 6 months ago

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@Sudoku Subbu – We can't say "concept" when we talk about "Fundamental theorem of algebra".

Andi Popescu - 4 years, 5 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 - 4 years, 6 months ago

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

Ricardo Adán Cortés Martín - 4 years, 6 months ago

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boss what r u trying to tel??

SARAN .P.S - 3 years, 10 months ago

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