Physics Question: If sin(theta)=theta, what does cos(theta)= ? too?
925769
sin(θ) ≈ θ\nSo cos (θ) ≈ 1  because cos (0) = 1\nMore accurately cos (θ) ≈ 1 - θ²/2
1046178
I think...\n\n# means theta\n\nsin(90-#)*cos(90-#)=tan#/sec^2#\ncos#sin#                =(sin#/cos#)/(1/cos^2#)\ncos#sin#                =(sin#cos^2#)/cos#\ncos#sin#                =cos#sin#\n\nUsing: sin(90-#)=cos#\n           cos(90-#)=sin#\n           tan#= (sin#/cos#)\n           sec^2#=(1/cos^2#)\n\nThat work?
668168
sin = sine\ncos = cosine\n\nthe sine of an angle in a right triangle is the side opposite divided by the hypoteneuse\n\nthe cosine is the adjacent side over the hypoteneuse.\n\nYou can look these values up in a table, or get them from some calculators.  google "sine" and you'll probably find a table.
890202
in trigonomertic form\n\nz = r cos t + ir sin t = 1 + i\n\nso r cos t = 1\n r sin t = 1\n\nsquare and add r^2 = 2\nso r = sqrt(2)\n\nnow cos t = 1/sqrt(2) and sint = 1/sqrt(2)\n\nso t = pi/4 (as cos pi/4 = 1/sqrt(2) and sin is also positive so 1st quadrant)\n\nso t = pi/4 and r = sqrt(2)
290580
It is the same as the graph of -cos(2 theta).\n\nIt starts at (0, -1) , has an amplitude of 1, and has a period of pi radians which means it passes through (pi/2, 1)\n\nOr in degrees, it has a period of 180 and passes through (90, 1)
899353
1) you basically need to draw out a triangle.  cot-1(x) returns the angle whose cotangent is equal to x, so let theta=cot-1(5/12).  this means cot(theta)=5/12, or tan(theta)=12/5.  the triangle associated with this problem is a 5-12-13 right triangle, so the sine of theta is just 12/13.\n2) break the velocities down into x-components and y-components, using the formula vx=vcos(theta) and vy=vsin(theta):\nvx=v1cos(theta1) +v2cos(theta2)\n  =580cos(332) +60cos(45)\n  =512+42\n  =555 mph\nvy=v1sin(theta1) +v2sin(theta2)\n  =-272+42\n  =-230 mph\nnow, use the pythagorean theorem to find the magnitude of the plane's velocity:\nv=sqrt(vx^2+vy^2)\n  =sqrt(307510+52838)\n  =sqrt(360349)\n  =600 mph\nnow, use the formula tan(theta)=vy/vx to find the direction:\ntheta=arctan(555/-230)\n  =-22.5 degrees\n  =337 degrees
437135
a cross b is a*b*sin theta (the angle between them)\nIn your case a and b are perpendicular theta=90 sin theta=1\na x b = 5x2x1 = 10 c where c is a vector perpendicular to their plane but the problem says |a x b| and this modulus means the algebraic answer only without vectors hence the answer is 10\nI hope this helps
730760
θ = α + β  where α = tan^-1 (1/3) and β = tan^-1 (1/2)\nie tanα = 1/3 and tanβ = 1/2\n\ntanθ = tan(α + β)\n= (tanα + tanβ)/(1 - tanαtanβ)\n= (1/3 + 1/2)/(1 - 1/3 * 1/2)\n= (5/6)/(5/6)\n= 1\nSo θ = ....., π/4, 5π/4, .....\nSo cosθ = ±1/√2 = ±√2/2
784184
θ =0.3t² rad\nω = dθ/dt = 0.6t rad/s\nα = d²θ/dt² = 0.6 rad/s²\n\nWhen t = 5\na. ω = 0.6t\n= 0.6*5 = 3 rad/s\n\nb. v = rω\n= 10*3 = 30m/s\n\nc. α = 0.6 rad/s²\n\nd a = v²/r (= rω² = 10*3² = 90 m/s²)\n= 30²/10\n= 90 m/s²
932480
sin(A+B)sin(A-B)/(cos²Acos²B)\n= (sinAcosB + cosAsinB)(sinAcosB - cosAsinB)/(cos²Acos²B)\n= (sin²Acos²B - cos²Asin²B)/(cos²Acos²B) (since (x + y)(x - y) = x² - y²)\n= sin²A/cos²A - sin²B/cos²B\n= tan²A - tan²B ← Option c)\n\nPlease note: sin(A + B) = sinAcosB + cosAsinB\nand sin(A - B) = sinAcosB - cosAsinB
324968
its just triangle mathe. remember soh -cah -toa... sine =opposite side/hypotenuse,  cosine=adjacent side/hypotenuse, tangent= opposite side/adjacent side....cosecant(hypotenuse/opposite side) is the opposite of sine, secant(hypotenuse/adjacent side) is the opposite of cosine and cotangent(adjacent side/opposite side) is the opposite of tangent
804214
sine 180=0\ncos 180= 1\ntan 180= 0\nsec 180= 1
958733
sin²y + cos²y = 1\nSo cos²y = 1 - sin²y\n= (1 - siny)(1 + siny)\nTherefore\ncos²y/(1 + siny) = 1 - siny .................... QED
184556
Let x(t) = A*cos( w*t + theta )\n\nthen x(0) = A*cos( theta )\nand dx/dt (0) = -A*w*sin(theta)\n\nso if you know x and its derivative in t=0 u have two equations and two unknowns ( A and theta ) . the rest is simple algebra .
721751
let theta be represented by 't' and integral sign be replaced by I. Then\n\nI[1 - cos(2t)dt\n= I[1*dt - cos(2t)*dt]\n= I(dt) - I[cos(2t)]dt\n= t - [sin(2t)]/2 + C ___________(1)\nwhere C = arbitrary constant of integration\n\nFrom (1)\nWhen t = 0, I(0) = 0 - 0 + C = C \nWhen t = T, I(T) = T - [sin(2T)]/2 + C\n\nFrom the Fundamental Theorem of Integral Calculus we can write that \nthe reqd. integral = I(T) - I(0)\n= T - [sin(2T)]/2 + C - C\n= T - [sin(2T)]/2\n= T - {sin(T)}*{cos(T)} [since sin(2T) = 2*{sin(T)}*{cos(T)}]
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.
523009
sine is positive and sec is negative so that means that theta is in the second quadrant.\ncosine and tangent are bot negative.\ncos(theta) = -sqrt(3)/2\ntan(theta) = -1/sqrt(3) = -sqrt(3)/3