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How do you calculate sec in trigonometry?
sec(x)=1/cos(x) - (hint: look at the third letter: sec->(1/)cos, cosec->(1/)sin, cot->(1/)tan)
Can two non-zero vector got a zero resultant?
Yes.If the angle between them is 90 degrees. As we know that A.B=|A| |B| cos (phi). When phi=90 degree,cos 90=0. Hence A.B= |A| |B| *0 =0.
How is the golden ratio worked out?
(a+b)/a=a/b=phi (the golden ratio, as defined) (a+b)/a=phi (we'll solve this equation) 1+b/a=phi (just changing the form of the left side a little) 1+1/phi=phi (a/b=phi so b/a=1/phi) phi+1=phi2 (multiply both sides by phi) phi2-phi-1=0 (rearrange) From here, we can use the quadratic equation to find the positive solution: phi=(-b+√(b2-4ac))/(2a) phi=(1+√(1+4))/2 phi=(1+√5)/2≈1.618
How do you prove that (1 plus cotx)2-2cotx 1(1-cos)(1 plus cos)?
Manipulate normally, noting:cot x = cos x / sin xcos² x + sin² x = 1 → sin²x = 1 - cos² xa² - b² = (a + b)(a - b)1 = 1²ab = baa/(bc) = a/b/c(1 + cot x)² - 2 cot x = 1² + 2 cot x + cot² x - 2 cot x= 1 + cot² x= 1 + (cos x / sin x)²= 1 + cos² x / sin² x= 1 + cos² x / (1 - cos² x)= ((1 - cos² x) + cos² x)/(1 - cos² x)= 1/(1² - cos² x)= 1/((1 + cos x)(1 - cos x))= 1/(1 - cos x)/(1 + cos x)QED.
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
