(sec x - 1) (sec x + 1) ; tan^2 x reduce the first expression to the second.?
(sec x - 1) (sec x + 1) = (sec² x - 1) -------------(a + b)(a - b) = (a² - b²) but sin² x + cos² x = 1 divide both sides by cos² x tan² x + 1 = sec² x so sec² x - 1 = tan² x so (sec x - 1) (sec x + 1) = tan² x
Reduce to a single trigonometric ratio?
[(1-cosx)(1-cos^2x)] simplifies to (1-cos^2x)/(cosx) . Since you already have a common denominator, you can merge (1/cosx) - (1-cos^2x)/(cosx) into a single term, giving you (1-1+cos^2x)/(cosx) (Don't forget to distribute the minus sign to both 1 and cosx) This simplifies to (cos^2x)/(cosx), which is the same thing as just cosx. Hurrah!
Power Reducing Formula in Trigonometry?
... ( sin⁴ x )( cos⁴ x ) = (1/16)( 16 sin⁴ x cos⁴ x ) = (1/16)( 2 sin x cos x )⁴ = (1/16)( sin 2x )⁴ = (1/16) sin⁴ 2x = (1/16). (1/4) ( 4 sin⁴ 2x ) = (1/64). ( 2 sin² 2x )² = (1/64). ( 1 - cos 4x)² = (1/64). [ 1 - 2 cos 4x + cos² 4x ] = (1/64). { 1 - 2 cos 4x + (1/2)( 1 + cos 8x ) } = (1/64). (1/2){ 2 - 4 cos 4x + 1 + cos 8x ) = (1/128)( 3 - 4 cos 4x + cos 8x ) ............................ Ans. ________________________________
Do trigonometric functions only work for right triangles?
The general rules we study are like:Sin X = P/HCos X = B/HTan X = P/BThese are special cases when we fix one angle as 90 .The rules like Sine rule and Cosine rule works on all cases.Fixing one angle as 90 and other as Angle X will make the measures of angles as 90,X and 90-X and this makes it congruent with any other right angle triangle with angle X as one of the angle by AAA Theorem. By congruent properties the ratio of sides will always be fixed and defined.
What should I do if I am not able to solve trigonometric problems of proving?
I am generalizing the answer for every domain.I had the same problem when I was first introduced to trigonometry. I was not able to solve even simple problems. What I did was but now follow in every domain:-Revise all the formulas,concepts and equations of trigonometry.Wrote all formulas and equations on a piece of paper.Every question I solved, I used to remember its that can help in solving future problems.Start from the basic problems.Whenever you can’t think how to proceed ahead, just think how you can reduce this by applying some identity or tricks.Remember all the tricks you ever used. It is very helpful in dealing problems.At least give half an hour to every problem if you can’t solve it. After that you can check the solution on internet, books,etc.Lastly but modt important . Practice and practice and practice……………….This is how you can improve your problem solving in every domain
Use the power reducing identities to rewrite the expression to one that contains a single trigonometric..?
2sin a cos a = sin 2a 2cos^2 a - 1 = cos 2a 4sin6xcos6x(2cos^2 6x-1) = (2sin 12x)(cos 12x) = sin 24x
Trigonometric.... lowering powers, cos^6 (x)?
You have known the formulas, just applied them. cos^6 x = (cos^2 x)^3 = ((1 + cos 2x) / (2))^3 = (1 + cos 2x)^3 / 2^3 = (1 + 3.cos 2x + 3.cos^2 2x + cos^3 2x)/8 = ( 1 + 3.cos^2 2x + (cos 2x)(3 + cos^2 2x) ) / 8 = ( 1 + 3(1 + cos 4x)/2 + (cos 2x)(3 + (1 + cos 4x)/2)) / 8 = (1 + 3/2 + (3cos 4x)/2 + (cos 2x)(3.5 + (cos 4x)/2) /8 = (1 + 3/2 + (3cos 4x)/2 + 3.5cos 2x + (cos 4x . cos 2x)/2 )/8 = (1 + 3/2 + (3cos 4x)/2 + 3.5cos 2x + (cos 6x + cos 2x)/4) /8 = (4 + 6 + 6cos 4x + 14cos 2x + cos 6x + cos 2x)/32 = (10 + 15 cos 2x + 6 cos 4x + cos 6x)/32 And you're done. Here we used the identity: cos A . cos B = (cos (A+B) + cos(A-B))/2 Hope that helps.
Use the power-reducing formulas to rewrite the expression 5sin^4 x as an equivalent expression?
with Euler's formula : sin^4(x)=1/8(cos(4x)-4cos(2x)+3)
Use the power-reducing formula to rewrite the expression?
6sin^4x and (sin^2 x) (cos^2 x) I have a whole section of homework on double-angle, power-reducing and half-angle formulas. I'm in tears over here because there's not a single problem that I can do! Please help :(