WEBVTT
00:00:00.800 --> 00:00:12.680
Solve root two sin 𝜃 plus root three cos 𝜃 equals two, where 𝜃 is greater than zero and less than or equal to two 𝜋.
00:00:13.820 --> 00:00:18.510
Give your answer in radians to two decimal places.
00:00:19.610 --> 00:00:28.630
In this question, we have a harmonic equation that we can solve by firstly recalling one of the addition rules of trigonometry.
00:00:29.790 --> 00:00:38.670
This states that sin of 𝐴 plus 𝐵 is equal to sin 𝐴 cos 𝐵 plus cos 𝐴 sin 𝐵.
00:00:39.690 --> 00:00:52.550
We begin by writing any harmonic equation in the form 𝑎 sin 𝜃 plus 𝑏 cos 𝜃 as 𝑅 multiplied by sin of 𝜃 plus 𝛼.
00:00:53.580 --> 00:01:13.250
In this question, we want to rewrite root two sin 𝜃 plus root three cos 𝜃 in the form 𝑅 multiplied by sin of 𝜃 plus 𝛼, where the values of 𝑅 and 𝛼 can be determined using the constants 𝑎 and 𝑏.
00:01:14.350 --> 00:01:26.580
We may simply wish to recall that 𝑅 is equal to the square root of 𝑎 squared plus 𝑏 squared and 𝛼 is equal to the inverse tan of 𝑏 over 𝑎.
00:01:27.470 --> 00:01:34.160
Alternatively, we can rewrite the right-hand side of our equation using the addition formula.
00:01:35.190 --> 00:01:46.630
After distributing the parentheses or expanding the brackets, we have 𝑅 sin 𝜃 cos 𝛼 plus 𝑅 cos 𝜃 sin 𝛼.
00:01:47.590 --> 00:02:04.030
We can then compare the coefficients of sin 𝜃 and cos 𝜃 on each side of the equation, giving us root two is equal to 𝑅 cos 𝛼 and root three is equal to 𝑅 sin 𝛼.
00:02:05.100 --> 00:02:12.620
We now have a pair of simultaneous equations that we can use to find the values of 𝛼 and 𝑅.
00:02:13.720 --> 00:02:24.630
Dividing equation two by equation one, we have root three over root two is equal to 𝑅 sin 𝛼 over 𝑅 cos 𝛼.
00:02:25.530 --> 00:02:34.870
Since sin 𝛼 over cos 𝛼 is equal to tan 𝛼, we have tan 𝛼 is equal to root three over root two.
00:02:35.870 --> 00:02:46.410
We can then take the inverse tangent of both sides of this equation such that 𝛼 is equal to the inverse tan of root three over root two.
00:02:46.540 --> 00:02:54.670
We can therefore see that 𝛼 is equal to the inverse tan of 𝑏 over 𝑎.
00:02:55.360 --> 00:03:05.570
Typing this into our calculator, ensuring it is in radian mode, we have 𝛼 is equal to 0.886 and so on.
00:03:06.410 --> 00:03:14.890
We can then substitute this exact value of 𝛼 back into equation one or two to calculate 𝑅.
00:03:16.000 --> 00:03:24.370
In equation two, we have 𝑅 is equal to root three over sin 𝛼, which is equal to root five.
00:03:25.390 --> 00:03:39.300
We could also have found this value of 𝑅 using our knowledge of the Pythagorean trigonometric identity, which leads us to the fact that 𝑅 is equal to the square root of 𝑎 squared plus 𝑏 squared.
00:03:40.250 --> 00:03:47.190
The square root of root two squared plus root three squared is equal to root five.
00:03:48.290 --> 00:04:01.000
Clearing some space, we now have the equation root five multiplied by sin of 𝜃 plus 0.886 and so on is equal to two.
00:04:01.960 --> 00:04:10.090
And we need to solve this equation for values of 𝜃 greater than zero and less than or equal to two 𝜋.
00:04:11.090 --> 00:04:16.020
We begin by dividing both sides of our equation by root five.
00:04:16.880 --> 00:04:24.530
This gives us the sin of 𝜃 plus 0.886 and so on is equal to two over root five.
00:04:24.530 --> 00:04:30.420
Next, we take the inverse sine of both sides.
00:04:30.850 --> 00:04:37.740
𝜃 plus 0.886 is equal to the inverse sin of two over root five.
00:04:37.740 --> 00:04:49.320
Ensuring that our calculator is in radian mode once again, the right-hand side becomes 1.107 and so on.
00:04:50.280 --> 00:04:59.370
We can now find one solution to our equation by subtracting 0.886 and so on from both sides.
00:05:00.150 --> 00:05:04.690
This gives us 0.221 and so on.
00:05:05.530 --> 00:05:13.640
Rounding this to two decimal places as required, we have 𝜃 is equal to 0.22.
00:05:14.650 --> 00:05:17.390
At this stage, we might think we are finished.
00:05:17.790 --> 00:05:23.680
However, we want all the values of 𝜃 that lie between zero and two 𝜋.
00:05:24.660 --> 00:05:29.840
One way of checking for any other solutions is using our CAST diagram.
00:05:30.740 --> 00:05:38.350
The positive angles between zero and two 𝜋 are measured in a counterclockwise direction as shown.
00:05:39.260 --> 00:05:46.910
Since the sine of our angle is positive, we know there will be solutions in the first and second quadrants.
00:05:47.800 --> 00:06:03.050
And from the symmetry of the sine function, we have a second solution where 𝜃 plus 0.886 and so on is equal to 𝜋 minus 1.107 and so on.
00:06:04.190 --> 00:06:14.630
We can then subtract 0.886 from both sides, giving us 𝜃 is equal to 1.148 and so on.
00:06:15.710 --> 00:06:23.540
Once again, rounding to two decimal places, we have 𝜃 is equal to 1.15.
00:06:24.620 --> 00:06:47.610
The solutions to the equation root two sin 𝜃 plus root three cos 𝜃 equals two, where 𝜃 is greater than zero and less than or equal to two 𝜋 radians, to two decimal places, are 𝜃 is equal to 0.22 and 1.15.