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Answer:
Why is cos0=1 and sin0=0?
⑴ Just imagine we have a right-angle ⊿ABC with the ∠C=90°, ∠A= 10° and∠B= 80°.
Take BC as the base, then we have a very steep slope with apex A.
For angle A=10°,the adjacent side would be ALMOST AS LONG AS the hypotenuse,ie
cos10° is close to 1.
As we gradually decrease angle A or as ∠A→0,cosA→1.
Eventually, cos A becomes 1 ,as A becomes 0→cos0=1
⑵imagine BC=1 ,AC=1,000 AB=√1.000,001
∠A would be ver small, SinA=1/√1,000,001≈0
As we increase AC say to 1,000,000,keeping BC=1,
AB becomes √(1,000,000,000,001),∠A→0, sinA=1/√1,000,000,001≈0
Eventually, it implies that sin 0=0
⑶EULERR’S FORMULA,
e^ix=cosx+isinx ,plug in x=0 and equating real and imaginary parts,
e^0=1=cos0+isin0 ,
cos0=1, sin0=0
⑷ POWER SERIES METHOD:
cosx=1—½x²+x^4/4!—x^6/6! +......
sinx =x —x³/3!+x^5/5!—x ^7/7!+...
plug in x=0
→sin0=0 , ‘cos 0=1
Step-by-step explanation:
hope my answer will help you
