what is : sin(2a) and cos(2a) ? and if there is many expressions give them all thanks in advance?
820147
sin(2a)=2*sin(a)*cos(a)\ncos(2a)=cos^2(a)-sin^2(a)
814944
we can  Use the trig identity\nsin(a+b)=sin(a)cos(b) + sin(b)cos(a) \nand we have\n\n10[sin(w.t)cos(π/4) + sin(π/4)cos(w.t)] \nwhich can be simplified as\n 10[√(2)/2sin(w.t) + √(2)/2cos(w.t)]\n= 5(√2)[sin(w.t) + cos(w.t)] \n\n a=5√2\n b=5√2
12429
Here are a couple i learned for double angle formulas.\n\nFor sin(a+b)= sin(a)cos(b)+cos(a)sin(b)\nSay out loud: "sine cosine cosine sine"\n\nFor cos(a+b)= cos(a)cos(b)-sin(a)sin(b)\nSay out loud: "cosine cosine SIGN sine sine"\nNotice the word "SIGN". The word SIGN is there to remind you that the SIGN of the rightmost term has the opposite SIGN of the input to the cosine on the left hand side of the equation (if its a+b, then subtract. if its a-b, then add).\n\nJust say it out loud like 3 times. It has a really nice rythm:\n\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n"Sine Cosine Cosine Sine"\n"Cosine Cosine SIGN Sine Sine"\n\nSee? Easy.
282780
1.sin(2x)=2sin(x)cos(x)\n2.tan(x)=sin(x)/cos(x)\n3.cos(x)=(1-sin^{2}(x))^(1/2)\n\nwe can use (3) in (2) and (1) to eliminate cos(x). Then substitute sin(x)=y:\n\n1.sin(2x)=2y.(1-y^{2})^(1/2)\n2.tan(x)= y/(1-y^{2})^(1/2)\n\ntherefore,\n\nf(y)=2y(1-y^{2})^{1/2}\n     +y/(1-y^{2})^{1/2}\n\nI hope that's what you wanted.
329959
1-CosA = 2 sin^2 A/2     -----1\n\nFirst whe have to find the value of the cos \n By Cos A = Root(1-Sin^2A)\nCos A = 5/13\nAnd \nThe apply the formaula of the 1\n1- 5/13 = 2Sin^2 A/2\nSinA/2 = Root(8/26) it gives the answer\n And the value of the cos A/2\nIs get by equation 1+ cos A = 2 Cos ^2 A/2
433978
Simply equate the functions as it asks you to;\nso you get 1-sinx = 2sin2x\nThen use the trig identity sin2x=2sinxcosx\nso you get 1-sinx=4sinxcosx\n\nCan you solve it from there?\nI'd next use cosx = sqrt (1-(sinx)^2)\nThen square the equation and solve it for x as you've probably done a million times before.\n\nYou'll probably get a quadratic equation in sinx. Use the quadratic formula to get some values out, and remember that you have to give answers between -360 and 360 degrees.
880753
we have cot A/2 \n\n= (cos A/2)/(sin A/2) = 2 cos^2A/2/sin A = (cos A+1)/ Sin A\n\nwe need to prove cosA(cos A +1)/sin A + (cos C+1)(cos C)/ sin C \n= 2 cos B(1+ cos B)/sin B\n\n1st term =  cos^2 A/sin A + cos A/sin A = (1-sin ^2A)/sin A+ cos A/sin A\n= (1+ cos A)/sin A - sin A\nas sin A sin B and sin C are in AP we need to show that\n\n(1+cos A)/Sin A, (1+ cos B)/sin B, (1+ cos C)/sin C are in AP\n\n1+ cos A = 1+ (b^2+c^2-a^2)/2bc\n= (2bc+b^2+c^2-a^2)/2bc\nso (1+cos A)/sin A = (2bc +( b^2+c^2-a^2)/2abc k\n\nas sin A/ a= sin B/b = sin C/c = k\n(1+Cos C)/sin C = (2ab+(b^+a^2-c^2)/2abck\n\nBy additing you should be able to get
962866
Anami's answer was correct and very simple.  You should give him credit for it.\n\nHere it is:\n\nID: cos(A+B)-cos(A-B)= -2sinAsinB\nso RHS: A=3x and B=x\n=> -2sin3xsinx = cos4x -cos 2x\nproved.\n\nWhat he did is this (only working with the left side):\n\ncos4x - cos2x\n------------------\n2sin3x\n\ncos(3x + x) - cos(3x - x)\n------------------------------\n2sin3x\n\ncos3x cosx - sin3x sinx - (cos3x cosx - sin3x sinx)\n-----------------------------------------------------------------\n2sin3x\n\n-2sin3x sinx\n---------------\n2sin3x\n\n-sinx\n\nis that more clear?
1092612
In quadrant one, sine, cosine, and tangent are positive. In quadrant two, sine is positive, and cosine and tangent are negative. In quadrant three, tangent is positive, and sine and cosine are negative. And in quadrant four, cosine is positive, and sine and tangent are negative.\n\nQuadrant 1 – Sin(+) Cos(+) Tan(+)\nQuadrant 2 – Sin(+) Cos(–) Tan(–)\nQuadrant 3 – Sin(–) Cos(–) Tan(+)\nQuadrant 4 – Sin(–) Cos(+) Tan(–)
419071
You can cheat and use a calculator to get the values, but I don't think this is what your teacher is looking for. \n\nFirst, use the relationship sin^2(x)+cos^2(x)=1 to find |sin(a)| and |cos(b)|. Pay close attention to the ranges of a and b to get the sign right-- sin(a) and cos(b) should both be negative. Doing this you should find:\n\nsin(a) = -12/13 \ncos(b) = (-2 sqrt(2))/3\n\nHere are two useful identities:\n\ncos(a-b) = cos(a)cos(b) + sin(a)sin(b)\ncos(2b) = 1 - 2sin^2(b)\n\nSince you now know cos(a),cos(b),sin(a) and sin(b), you can find that:\n\ncos(a-b)=(12-10sqrt(2))/39\ncos(2b)=7/9
835259
I'll give you a hint and let you try to figure out the rest yourself (unless someone else tells you the exact answer). =)\n\nUse your trig identities:\n\ntan A = sinA/cosA\n\ncot A = 1/tanA or cot A = cosA/sinA
1122517
sin^2+cos^2=1\nsin=1/cosec\nsin=opp side/hypotheses side\ncos=adj side/hypotheses side\nhyp^2=opp^2+adj^2\n\nSin*cosec*cos=cos
278456
So you know the ones for sin(2a) and cos(2a). Cool.\n\nNow, tan(2a) is the same as sin(2a)/cos(2a), since tan = sin/cos (definition of tangent).\n\nSub in the formulae for sin(2a) and cos(2a) :\n\n2sin(a)cos(a)\n----------------\ncos(a)^2-sin(a)^2\n\nDivide both numerator and denominator by (cos(a))^2\n\n2tan(a)\n----------\n1-tan(a)^2
1054686
Job one, convert everything to trig functions of a single angle, namely x.\n\nsin2x=2sinxcosx\ncos2x=cos^2(x)-sin^2(x)=2cos^2(x)-1=1-2sin^2(x)\n\nThen substitute in and simplify:\n2sinxcosx+2(2cos^2(x)-1)=1+sinx-4cosx\n2sinxcosx+4cos^2(x)-2-1-sinx+4cosx=0\n4cos^2(x)+2sinxcosx-sinx+4cosx-3=0\n\nFactor this into:\n(2cosx-1)(2cosx+sinx+3)=0\n\nset each parantheses equal to zero\n2cosx-1=0   \ncosx=1/2\nx=arccos(1/2)\nx={pi/3, 5pi/3} + 2pi*n  where n=0,1,2....\n\nAnd\n\n2cosx+sinx+3=0\n\nSolved numerically gives imaginary solutions of\n{-i*ln((-2/5)-(i/5),-i*ln(-2-i)}\n\nSolved....
788868
cosθ = 2cos²θ/2 - 1\nSo cos²θ/2 = ½(1 + cosθ)\nSo cos²(π/12) = ½(1 + cos(π/6))\n= ½ + √3/4\n= ¼(2 + √3)    \nSo cos(π/12) = + √{¼(2 + √3)}\n= ½√(2 + √3)\n\nLet √(2 + √3) = a + b√3\nSquare both sides\n2 + √3 = a² + 3b² +2ab√3\n\nSo a² + 3b² = 2 and 2ab = 1\na = 1/√2 and b = 1/√2 \nSo √(2 + √3) = ½(√2 + √6)\n\nSo cos π/12 = ¼(√2 + √6)
900488
cos(x) +sin(x)tan(x)=sec(x)\nmultiply by cos(x) on both sides\ncos^2(x)+sin^2(x)=1\n1=1\nI believe they are probably asking you to prove the identity
710344
C - 3 possible answers\n\nIn any given right triangles with the two other angles being a and b,\n\nsin a = cos b AND sin b = cos a,\n\n\nas such, the first and fourth combination will give one answer, the other two will either give square trigonometric values of each angle i.e. sin^2 a and cos^2 a  OR sin^2 b and cos^2 b.\n\nThe only condition that will not satisfy this is when the angle is 45 degrees, which means a = b and contradicting the given statement of a being less than b.\n\nThus the answer is C.