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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Note: cos2θ=1⇒cosθ=1. Do you see why?
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why staff iam sending the sqare to that side so cosθ=1 hmmm.. ttooo simple concept
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If x2=1, is x=1 the only solution?
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x2=1 then x may be -1 or +1 thanks calvin i'll change
ya ya i understand your concept ifLog in to reply
The best approach would be to say 0=x2−1=(x−1)(x+1) and hence x=1,−1.
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Another thing... you don't just consider the first quadrant angle.. Consider also its coterminal angles..
Remember the Pythagorean Identity: sin2θ+cos2θ=1 If cos2θ=1, this implies that: sin2θ=0sinθ=0 ¿For what angles is this true? For any angle of the form θ=nπ As can be easily verified seeing any graph of the sine function.
boss what r u trying to tel??